pmb
Oct24-03, 05:36 PM
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
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