JS Atomic Physics Mathematica Lab
By David Long, 03457885
Exercise 1:
The aim of exercise 1 was to consider gyroscopic motion in a gravitational field, as a
prelude to examining angular momentum in a magnetic field. This was done by first
trying to find the Larmor Frequency of precession, given the diagram of forces, and
the angular momentum of the gyroscope in SI units.
Angular
momentum
L=I
Angular
velocity 
r
F
L = 0.05 (sincos(t), sinsin(t), cos). As r is parallel to L, and the distance from
the ground to the centre of mass of the gyroscope is 0.05m, therefore r is also equal to
L
 r  F , therefore, the equation is
0.05 (sincos(t), sinsin(t), cos). Given
t
as follows:



L
 0.05 sin  sin( t ) i  0.05 sin  sin( t ) j  0 k
t
rF 



i
j
0.05 sincos(t) 0.05sinsin(t)
0
-0.5

k
0.05cos
0

= 0.025 cos  i  0.025 sin  cos(t ) k
 0.05 sin  sin( t ) = 0.025 cos
 0.05 sin  sin( t ) = 0
0 = 0.025 sin  cos( t )
From these equations, it may easily be seen that we get Ω = 0.5.
The next task was to find a value for the Angular velocity ω assuming g=10m/s/s. As
L=Iω, therefore ω=L/I. The angle θ is 45o, as the radius of the gyroscope, and the
distance to the centre of mass are the same (0.05m). From this then, we get a value of
L=0.05. I on the other hand is equal to mr2, for a hoop of radius r, giving I =0.000125.
These results give a value of ω=400, which is the angular velocity of the gyroscope.
Exercise 2:
(a): In the first part of exercise 2, the aim was to show that the Larmor frequency
Ω = γB. This was done as follows:

1 d
Given: 
   B and    (sin  cos t , sin  sin t , cos  , B  B k .
 dt
B=



i
j
 sin  sin t
0
k
 cos
B
 sin  cos t
0


= Bμsinθsin(Ωt) i - Bμsinθcos(Ωt) j


d
= -μΩsinθsinΩt i + μΩsinθcosΩt j
dt
Therefore: 
1

(-μΩsinθsinΩt) = BμsinθsinΩt,
Giving: Ω = γB
(b)
Q.E.D.
mu x
2
1.5
1
0.5
time
-0.5
1
2
3
4
5
6
-1
-1.5
-2
In the above graph, the magnetic moment B has been set at B = √ 3. The period may
be seen to be roughly 3.6.
2Pi/T = γB, therefore, 2Pi/3.6 = √ 3 (γ = 1)
1.74 = 1.74. Q.E.D.
(c) The effects of the magnetic moment precession in a static magnetic field were then
graphed as an animated file:
In the graphs shown below, the settings were:
b1 = 1.; b2 = 1.; b3 = 1; l10 = 1; l20 = 0; l30 = 1; In these cases, bx denotes the
magnetic field strength in the given directions, while lx gives the initial magnetic
moments in the given directions. The blue line denotes the magnetic field direction,
while the green line denotes the magnetic moment. As may be seen, for the given
readings of b & l, the magnetic moment precesses about the magnetic field direction.
In the graphs shown below, the settings were:
b1 = 0.; b2 = 0.; b3 = 1; l10 = 2; l20 = 2; l30 = 2; As can be seen, changing the b
values changes the strength and direction of the magnetic field, while changing the l
values changes the strength of the initial magnetic moments, and hence the
precession.
The final part of exercise 2 involved saving the results as a gif file which gives an
animated view of what happens when the input values are changed.
Exercise 3:
The third part of the experiment was concerned with the coupling of angular momenta,
namely the coupling of total orbital angular momentum, and total spin angular
momentum. (L-S coupling / Russell-Saunders Coupling). The first part involved the
dL
dS
 S  L;
 L  S . It was to be proved that the vector
differential equations:
dt
dt
sum of two coupled angular momentum remain fixed in space. This was done by
d
showing that: ( L  S )  0 .
dt
d
dL dS
(L  S ) =
+
dt
dt
dt
Therefore; S  L + L  S = 0
This implies: S  L = - L  S .
d
( L  S )  0 . Q.E.D.
Therefore,
dt
To do this,
The physical reason why this must occur is that the total torques of the coupled
angular momentum must equal 0, which occurs in this case.
In the second part of the exercise, a linear relationship was found between (Fig. 1) the
initial magnetic moment of the spin angular momentum and the precession frequency
(Fig. 2) the initial magnetic moment of the orbital angular momentum and the
precession frequency, and (Fig. 3) the coupling constant ξ and the precession
frequency. This Linear relationship shows that all three variables are proportional to
the Precession frequency.
Figure 1
2.5
2
Frequency
1.5
1
0.5
0
0
2
4
6
8
10
Magnitude of spin magnetic moment
It can be seen above that the magnitude of the spin magnetic moment is linearly
proportional to the precession frequency.
12
Figure 2
2.5
2
Frequency
1.5
1
0.5
0
0
2
4
6
8
10
12
Magnitude of Orbital magnetic moment
Again, the linear relationship between the magnitude of the orbital magnetic moment
may be seen from the graph above.
Figure 3
3
2.5
Frequency
2
1.5
1
0.5
0
0
1
2
3
4
5
6
Coupling Constant
Finally, the linear relationship between the Coupling constant ξ and the precession
frequency may be seen above.
7
Finally, given that two angular momenta have Cartesian coordinates (1, 0, 1) and (-1,
0, 1) at some moment in time, find a value for Ω (the precession frequency). This was
done as follows:
dL
dS
 S  L;
 L  S
dt
dt
L = (cos Ωt, sin Ωt, 1); S = (-cos Ωt, -sin Ωt, 1)
Therefore L S =



i
j
sin t
- sin t
k
1
1



i
j
- sin t
sin t
k
1
1
cos t
- cos t
Giving: (2sin Ωt, -2cos Ωt, 0)
While S  L =
- cos t
cos t
Giving: (-2sin Ωt, 2cos Ωt, 0).
dL
dS
= (Ωsin Ωt, -Ωcos Ωt, 0), and
= (-Ωsin Ωt, Ωcos Ωt, 0)
dt
dt
dL
dS
 S  L;
 L  S , and letting ξ = 1, it can be
dt
dt
seen that Ω = 2, and also that the points (1,0,1) and (-1,0,1) are indeed solutions to
the given equation. Q. E. D.
By inspection of the equation
Exercise 4:
In exercise 4, the effect of placing a pair of coupled angular momenta in a static
magnetic field was examined. In this case it was found that the sum of the coupled
angular momenta precessed about the direction of the static magnetic field, while both
angular momenta precessed about their own sum. This can be explained with more
clarity by examining the graphs of angular momentum components as functions of
time:
sx
2
jx
2
1.5
1.5
1
1
0.5
-0.5
-1
-1.5
-2
0.5
10
20
30
40
50
60
time
-0.5
-1
-1.5
-2
10
20
30
40
50
60
time
The two graphs above show effectively the same function. The graph on the left
shows the total angular momentum of the entire system, while the graph on the right
shows the angular momentum of the sum of the two coupled angular momenta. It can
be seen that the graph on the left mimics the one on the right, except that it is not as
smooth. This is due to the precession of the two coupled angular momenta about the
axis of their sum. Although the sum of the coupled angular momenta precesses
smoothly about the direction of the static magnetic field, the fact that the two coupled
angular momenta are precessing about the axis of their sum adds the uncertainty in
detail to the graph of the total angular momentum of the system.
The whole explanation may be more easily visible from the animation, snapshots of
which are reproduced below:
If the coupling between the magnetic field and the angular momenta were much
stronger than that between the angular momenta themselves, I would expect both the
total angular momentum and the sum angular momentum graphs to be very smooth,
as the effects of the coupling of the angular momenta themselves would no longer
affect the total angular momentum of the system:
sx
2
jx
2
1.5
1.5
1
1
0.5
0.5
-0.5
10
20
30
40
50
60
time
-0.5
-1
-1
-1.5
-1.5
-2
-2
10
20
30
40
50
60
time
This can be seen from the graphs shown above. The graph on the left shows the
angular momentum of the total system, while the graph on the right shows the angular
momentum of the sum of the two coupled angular momenta. As expected, they are
both smooth due to the lack of any effects of the coupling between the two angular
momenta. Both graphs above were produced by setting ξ = .00001, as opposed to ξ =
1 for the two original graphs shown above.
Exercise 5:
The first task in exercise 5 was to verify that the Mathematica code:
solution = NDSolve[{l1'[t]l3[t] (b2 + B2 Cos[ t]) - (b3 + B3 Cos[ t])l2[t],
l2'[t]l1[t](b3 + B3 Cos[ t]) - (b1 + B1 Cos[ t]) l3[t],l3'[t]l2[t] (b1 + B1 Cos[ t])
- (b2 + B2 Cos[ t])l1[t],l1[0]l10,l2[0]l20,l3[0]l30},{l1,l2,l3},{t,0,40
Pi},MaxSteps8000,MaxStepSize  0.05];
Corresponded to the equation: (
d
) lab  B tot  
dt
By inspection, this is the case.
mu z
1
0.75
0.5
0.25
time
-0.25
20
40
60
80
100
120
-0.5
-0.75
-1
The next part of the exercise involved
comparing the animated file (on the right)
and the graph of the z component of the
magnetic moment (on the left) and the
resonance frequency. As can be seen above,
at the resonance frequency, the motion of
the magnetic moment becomes smooth and regular, as opposed to any other frequency,
where the motion is jagged, and irregular.
mu z
1
0.75
0.5
0.25
time
-0.25
-0.5
-0.75
-1
20
40
60
80
100
120
As can been seem on the left, when the
Larmor frequency is set to 0.9 times the
resonance frequency, the amplitude of
the z-component of the magnetic
moment is greatly diminished. This is
due to the lack of resonance, and the
conflicting frequencies, which combine
to lower the amplitude and the period,
hence increasing the frequency.
The below graph (figure 4) shows the relationship of the amplitude of the magnetic
moment with increasing frequency. As can be seen, there is a peak at the frequency =
1, as this corresponds to the resonance frequency of the system. This graph kept the
magnetic field constant while the readings were taken.
Figure 4
2.5
2
Amplitude
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
Frequency
Figure 5
1.45
1.4
Amplitude
1.35
1.3
1.25
1.2
1.15
0
2
4
6
8
10
12
Magnitude of Magnetic Field
The above graph (Figure 5) shows the relationship between the amplitude and the
magnitude of the magnetic field for a fixed frequency. As can be seen, the amplitude
is fairly constant at about 1.35. This shows that the width of the resonance depends
explicitly on the magnitude of the time varying field.