- #1

- 523

- 0

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

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter jdstokes
- Start date

- #1

- 523

- 0

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

- #2

- 277

- 2

How is the relativistic momentum?

- #3

- 2,952

- 0

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

- 458

- 1

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

Share: