- #1

jdstokes

- 523

- 0

[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m} [/itex]?

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- Thread starter jdstokes
- Start date

- #1

jdstokes

- 523

- 0

[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m} [/itex]?

- #2

lzkelley

- 277

- 2

How is the relativistic momentum?

- #3

pmb_phy

- 2,952

- 1

Yes. But just to be clear, using your symbols, the transverse mass = gamma*m.

[itex]\gamma \frac{mv^2}{r} = qvB \implies \frac{v}{\sqrt{1-(v/c)^2}} = \frac{qBr}{m} [/itex]?

Pete

- #4

pam

- 458

- 1

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

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