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Larmor's Theorem
Larmor's Theorem: If a system of charged particles, all having the same q/m ratio, acted on by
their mutual (central) forces, and by a central force toward a common center, is subject to a weak,
uniform B, its possible motions will be the same as the motions it could perform without the B,
superposed upon a slow precession of the entire system about the center of force with angular
velocity:  = -(q/2m)B .
This theorem is based on the similarity of the Coriolis force and the magnetic force:
Fc = -2mv*,
Fm = qvB .
Proof: Let's consider this situation from the inertial frame:
mak = ΣFon k = FkC + ΣiFki + qvkB,
where FkC is the force on the kth particle due to the central force, and Fki is the force on the kth
particle due to the ith particle.
Let's look at the system from a precessing frame, one that is rotating about the central
force position with a frequency of . Recall from Eq. 10b in the Rotating Coordinate Systems
section:
(with A = r, dA/dt = v and d²A/dt² = a):
a = a* + (d/dtr) + ((r)) + (2v*) .

Therefore, Newton's Second Law becomes (with d/dt = 0, and dropping the subscript k):
ma = ma* + m((r)) + (2mv*) = FC + ΣFi + qvB .
Recall from Eq. 3c in the Rotating Coordinate Systems section:
d*A/dt = dA/dt + (A),
v* = v + (r),
or
or with A = r,
v = v* + (r) ,
Newton's Second Law becomes:
ma = ma* + m([r]) + (2mv*) = FC + ΣFi + (qv*B) + q([r]B) .
We now use the vector identity, (ab) = -(ba), twice:
ma = ma* + m([r]) - (2mv*) = FC + ΣFi + (qv*B) - q(B[r]) .
We now re-group terms so that terms that look alike are together:
ma* + ({m + qB} [r]) - (v*{2m + qB}) = FC + ΣFi .
Now if we let ω = -(q/2m)B as the theorem suggests, then the (v*{2m + qB }) term
goes to zero and the {m + qB} [r] term becomes +(q/2)B  [-(q/2m)Br]
so that the final expression in the rotating frame is:
ma* = FC + ΣFi + (q²/4m){B  [Br]}.
If the applied magnetic field is weak, then the last term is very small (being of the order of B²)
and can be neglected. This then says that the system acts in the rotating frame as if there were
NO magnetic force due to the external weak field! This is what Larmour's theorem says!
Cyclotron frequency: If we consider a charged particle moving in a magnetic field and moving 
to the field, then Newton's Second Law says: Fmagnetic qvB = ma , and since the a is  to v,
a  = acircular = v²/r , so we get qvB = mv²/r , or qBr = mv; but for circular motion, v =
r , so we get: cyclotron = qB/m, NOT -qB/2m as in the Larmour case.
QUESTION 1:
Why is the frequency of Larmour precession different than that of the cyclotron frequency?
QUESTION 2:
Does Larmour’s Theorem apply to the electrons in an atom subjected to a magnetic field of 1T?
(See next page for answers only after you have tried to answer them yourself.)
ANSWER 1:
The cyclotron frequency is ωcycltron = qB/m. The Larmour frequency is ωLarmour = qB/2m.
The Larmour frequency only applies if the centrifugal term associated with the rotation is much
smaller than the central or internal forces of the charged particles. Recall from the previous
page:
ma* = FC + ΣFi + (q²/4m){B  [Br]}. If the applied magnetic field is weak, then the
last term is very small (being of the order of B²) and can be neglected.
If FC + ΣFi is negligible, then we have ma* = (q²/4m){B  [Br]}.
Let’s first look at the direction of this “centrifugal” type term.
[Br] has to be in a direction perpendicular to B, so we only need to consider the components of
r that are perpendicular to B. For instance, let’s consider B in the z direction, and only consider
the x and y components of r. For the direction of the [Br] term we get: z× x = y and z× y = -x.
Now for the direction of the {B  [Br]} we get: z× y = -x and z× -x = -y so that the direction is
that of –r. This leads to ma* = (q²/4m){B  [Br]} becoming ma* = -(q²B2/4m)r where the
minus sign indicates that the acceleration is towards the center, but this is equivalent to a circular
acceleration of ω’2r, so we have mω’2r= q²B2r/4m , or ω’ = qB/2m .
This ω’ is in addition to the ωLarmour, so the total angular frequency is
ωcylotron = ωtotal = ωLarmour + ω’ = qB/2m + qB/2m = qB/m.
ANSWER 2:
Larmour’s theorem holds only if: (q²/4m){B  [Br]} « FC + ΣFi , so we need to look at the
magnitudes of the left and right sides for an electron circling a nucleus with an applied magnetic
field of 1 Tesla. For an atom, the radius is about 1 x 10-10 m, q for the electron and proton is 1.6
x 10-19 Coul., and the mass of the electron is 9.1 x 10-31 kg.
The maximum value of (q²/4m){B  [Br]} is
q2B2r/4m = (1.6x10-19C)2*(1T)2*(1x10-10)m / [4*9.1x10-31kg] = 7x10-19 Nt.
The force between the electron and the proton, FC is
kqeqp/r2 = (9x109Nt*m2/C2)* (1.6x10-19C)2 / (1x10-10m)2 = 1.4x10-8 Nt.
The “centrifugal” term is indeed much smaller than the central force!
For a magnetic field of 1 Tesla, the Larmour frequency for the electron would be:
ωLarmour = qB/2m = (1.6x10-19C)*(1T)/{2*9.1x10-31 kg) = 9x1010rad/sec, or
fLarmour = ωLarmour /2π = 1.5 GHz (in the microwave region).