Cyclotron formula and Transverse Mass

pmb

The cyclotron forumla is derived here
See -www.geocities.com/physics_world/cyclotron.htm

That's a lot of work for something so simple. Recall that the relativistic force can be written as
equation. Recall that

F = M_t A_t + A_L M_L

where

M_t = transverse mass
A_t = transverse acceleration

M_L = longitudinal mass
A_L = longitidinal acceleration

It can be shown that transverse mass = (relativistic) mass = m. For the cyclotron motion described in above link A_L = 0 and a = A_t = v^2/r. Therefore

F = dp/dt = M_t A_t = m A_t = ma

where a = transverse acceleration = v^2/r and p = mv = gamma*m_o*v. Therefore it follows that

F = ma = e[E + (v/c)xB]

So, in this particular instance, F = ma is 100% correct if m = relativistic mass and a is transverse acceleration. Plugging vxB = vB (B is now a magnitude rather than a vector)

ma = m(v^2/r) = vB

Canceling a factor of v and moving the r to the otherside gives

mv = vBr

p = m_o*v/sqrt[1-(v/c)^2] = mv

p = vBr

which is the relativistically correct relation for cyclotron motion.

Someone claims that this is not simpler than the normal derivation derived above [Then again the same people didn't know that F = q[E + vxB] either]. Does anyone know of a simpled/easier way to derive the cyclotron formula?

Pete

Related General Physics Workshop News on Phys.org

"Cyclotron formula and Transverse Mass"

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