Relativistic cyclotron frequency

  • Thread starter jdstokes
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  • #1
jdstokes
523
0
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]?
 

Answers and Replies

  • #2
lzkelley
277
2
How can you express the centripetal force in terms of momentum (with other terms)?
How is the relativistic momentum?
 
  • #3
pmb_phy
2,952
1
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
 
  • #4
pam
458
1
It's clearer in terms of momentum pv/r=qvB-->p=qBr.
 

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