# Special relativity, circular motion

1. Aug 16, 2011

1. The problem statement, all variables and given/known data
A charged particle (mass $m$, charge $q$) is moving with constant speed $v$. A magnetic field $\vec{B}$ is perpendicular to the velocity of the particle. Find the strength of the field required to hold the particle on a circular orbit of radius $R$.

2. Relevant equations
$\vec{F} = q\vec{v} \times \vec{B}$
$\vec{F} = m\vec{a}_c$

3. The attempt at a solution
Well, I know that in the "classical" case this is fairly easy. One just sets

$qvB = ma$,

and since $a = \frac{v^2}{R}$, one gets

$qvB = m \frac{v^2}{R}$
$\Rightarrow B = \frac{mv}{qR}$

However, I am not sure if I can use this here, because the particle is assumed to be travelling at close to the speed of light. I have read somewhere that I should use the relativistic mass in the calculation of the centripetal force, i.e.

$F = \frac{\gamma m v^2}{R}$,

but I am not sure why this is the case. Could anyone help?

Last edited: Aug 16, 2011
2. Aug 17, 2011

### mathfeel

One way to see this is go into the momentarily comoving frame (not an inertial frame, obviously) of the particle. In this frame, the particle is at rest so the Lorentz force only comes from the electric field in this frame, which is: $q E^{\prime} = \gamma q \mathbf{v} \times \mathbf{B}$
So it experiences a force which is perpendicular to both $\mathbf{v}$ and $\mathbf{B}$ with the $\gamma$ as promised.