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The Different Types of Particle Motion
Zoe Zontos
November 16, 2016
1 Introduction
The purpose of this project was to investigate different types of particle motion. The study
of the motion of particles in space physics is important and provides insight into understand-
ing how dynamical processes in plasmas work. It also provides insight on the microscopic
behavior of particles which is significant in order to understand plasma and space physics
on a macroscopic level [1]. In this project, I worked with four types of particle motion:
cyclotron motion, gradient-B drift motion, magnetic mirroring, and a magnetic dipole.
2 Cyclotron Motion
a) I first investigated the cyclotron motion of a particle. The analytical velocity vector of a
particle undergoing pure cyclotron was given as
v(t) = (v⊥ sin(ωct))ˆx + (v⊥ cos(ωct))ˆy + (v )ˆz. (1)
In order to determine the position of the particle as a function of time, I integrated v(t) to
obtain t
0
v(t)dt =
t
0
(v⊥ sin(ωct))ˆx + (v⊥ cos(ωct))ˆy + (v )ˆz)dt. (2)
From this integration, the position vector with respect to time for the particle was determined
to be
r(t) =
v⊥
ωc
(1 − cos(ωct))ˆx +
v⊥
ωc
sin(ωct)ˆy + v tˆz. (3)
Equation 3 represents the analytical solution for the position vector as a function of time
for a particle undergoing pure cyclotron motion. (NOTE: The equation ωc = |q|B
m
represents
the cyclotron frequency where q is the charge of the particle, B is the magnetic field value,
and m is the mass of the particle [2].)
b) I plotted the graph of the cyclotron motion of a proton showing the trajectories of the
analytical solution along with the numerical solution as shown in Figure 1. For the proton,
the mass is 1.672e-27 kg, the charge is 1.602e-19 C, and I used a magnetic field value of
45,000 nT.
1
Figure 1: Analytical and numerical solutions for the motion and position of a proton through
a uniform magnetic field.
The resulting plot revealed that the particle trajectories for the analytical and numerical
solutions were similar within a uniform magnetic field which was expected.
c) Utilizing the same technique as in part b, I repeated the process and plotted the analytical
and numerical solutions of cyclotron motion for an electron. For the electron, the mass is
9.109e-31 kg, the charge is the same as that of a proton, and the same magnetic field value
was also used.
2
Figure 2: Analytical and numerical solutions for the motion and position of a charged electron
within a uniform magnetic field.
The plot in Figure 2 revealed that, like the proton, the analytical and numerical solutions
were similar for the electron within a uniform magnetic field as well. Figures 1 and 2 both
showed that the direction of the proton and electron were clockwise and counterclockwise
(respectively) due to the Lorentz force F = q(E + v × B) acting on the particles where E =
0 and a constant B field for both, but the charges have opposite signs.
3 Gradient-B Drift
a) The next part of this project focused on the gradient-B drift motion of a proton. As
explained in Hughes, the gradient-B drift has an E=0 and a perpendicular gradient present
in the field strength. Wherever the magnetic field is changing, the particle will travel through
as long as it has some initial perpendicular velocity [2]. Introducing a magnetic field with a
gradient to the proton, I simulated and plotted the motion and position of the proton while
under the influence of gradient-B drift motion alongside with its cyclotron motion. The
equation for magnetic field was given as
B(x, y, z) = Bo(1 +
1
L
x)ˆz, (4)
3
where the reference magnetic field used was Bo = 45,000 nT, the initial velocity in the y-
direction was v(0) = 4000 m/s, and the L = 2 m. The magnetic field was kept strictly in the
z-direction while also including a linear variation in order to maintain a constant gradient
on the particle.
Figure 3: Motion and position of a proton while under the influence of gradient-B drift
motion and still undergoing cyclotron motion.
The plot of the proton in Figure 3 shows that the trajectories of its cyclotron motion and
gradient-B drift motion appear to be approximately equal.
b) In order to determine the position vector of the particle, I started by using the provided
equation for the magnetic field vector
B = Bo(1 +
1
2
x)ˆz, (5)
and the equation for the gradient-B drift which according to Hughes was given as
v B =
1
2
v2
⊥
ωc
B × B
B2
, (6)
where
B × B = −
B2
o
2
(1 +
1
2
x)ˆy. (7)
I plugged B × B into Equation 6 for v B which resulted in
4
v B =
mv2
⊥B2
o (1 + 1
2
x)
4qB3
o (1 + 1
2
x)3
ˆy. (8)
Simplifying the solution left the v B to be
v B =
mv2
⊥
(1 + 1
2
x)3
ˆy. (9)
I then integrated v B with respect to t and obtained an expression for the position r(t):
r(t) =
v⊥
ωc
[1 − cos(ωct)]ˆx + [
v⊥
ωc
sin(ωct) +
mv2
⊥
4qBo(1 + 1
2
)x2
t]ˆy + v tˆz. (10)
I plotted the analytical solution for r(t) which is seen in Figure 3 where the trajectories of
the particles are relatively the same [2].
c) In order to demonstrate how changing the initial velocity affects the the gradient-B drift
motion of the particle, I plotted the proton’s motion and position when its initial velocity
was doubled (left) and halved (right) as shown in Figure 4.
Figure 4: Motion and position of a proton while under the effects of gradient-B drift motion
when its initial velocity is doubled (left) and halved (right).
The results in Figure 4 revealed that changing the velocities of the particle affected the
particle’s Larmour radius as well as its y-direction component of position. The y-direction of
the particle’s position doubled when the velocity was doubled and halved when the particle’s
velocity was halved. In the case where the initial velocity of the proton was doubled, the
Larmour radius, rL = mv⊥
|q|B
= v⊥
ωc
, also doubled. Likewise, when the initial velocity was cut
in half, the Larmour radius also decreased by half.
4 Magnetic Mirrors
a, b) This part of the project focused on using a magnetic mirror to simulate a mirroring
’bottling’ motion undergone by the particle due to it moving through a region of a weak field
5
surrounded by a strong field. Using Gauss’s Law, I determined the radial component of the
magnetic field. Starting with the expanded form of Gauss’s Law in cylindrical coordinates:
• B =
1
ρ
∂
∂ρ
(ρBρ) +
∂Bφ
ρ∂φ
+
∂Bz
∂z
= 0, (11)
I first solved for Bρ:
∂
∂ρ
(ρBρ) = −ρ
∂Bz
∂z
, (12)
which simplified to
ρBρ = −
ρ2
2
∂Bz
∂z
, (13)
and even further
Bρ = −
ρ
2
∂Bz
∂z
. (14)
I then rewrote Equation 14 as
Bρ = − x2 + y2
Boz
a2
. (15)
Using the result found in Equation 14 and 15, I then converted to Cartesian coordinates
which resulted in a Bx(x, y, z) and By(x, y, z) shown below.
For Bx(x, y, z):
Bx(x, y, z) = Bρ(z)cosθ, (16)
which became
Bx(x, y, z) = −x
Boz
a2
. (17)
And for By(x, y, z):
By(x, y, z) = Bρ(z)sinθ, (18)
which became
By(x, y, z) = −y
Boz
a2
. (19)
The resulting magnetic field vector in Cartesian coordinates was
B(x, y, z) = −x
Boz
a2
− y
Boz
a2
. (20)
c) Choosing the origin as the initial position and using initial velocity components of vy0 =
vz0= 10,000 m/s, I plotted the path of the particle to demonstrate that it mirrors correctly
as shown in Figure 5.
6
Figure 5: Mirroring of the particle with an initial velocity of 10,000 m/s and its initial
position at the origin.
The mirroring of the particle in Figure 5 was correct; the particle moved in a path that was
’mirrored’ on both sides (it moved symmetrically).
Figure 6: The particle’s velocity versus time in the axis of the bottle (z-direction).
7
The particle did mirror correctly, and I estimated the period of mirror motion by looking
at the z-direction velocity of the particle inside the bottle in Figure 6. The bounce period
minima appeared to be at around 0.280s and 0.075 s, so the estimated bounce period is
(0.270 s) - (0.075 s) = 0.195 s which resulted in an estimated frequency of (1/0.195 s) = 5.13
Hz.
d) Using the equation for the mirror force, I solved for the the parallel equation of motion
for the specific magnetic field used.
The mirror force provided was written as
Fz = −µ
∂Bz
∂z
= m
d2
z
dt2
, (21)
where the magnetic moment µ =
1
2
mv2
⊥
B
, so, taking the second derivative, I determined
d2
z
dt2
= −
mv2
⊥
mBo
(
2Boz
a2
) = −v2
⊥
z
a2
. (22)
Solving for the differential equation yields:
z(t) = Asin(
v⊥t
a
) + Bcos(
v⊥t
a
). (23)
I applied the initial conditions of z(o) and z(t) to find that
z(0) = Bcos(
v⊥t
a
), (24)
and
z(t) = Asin(
v⊥t
a
). (25)
The resulting equation for the angular frequency can be written as
f =
v⊥
2πa
. (26)
Solving for the frequency and period, I obtained a value of 5.305 Hz for the frequency and
a period of t = 0.189 s. Comparing the calculated values to that of the estimated values,
there was a 3.17-3.29 % difference between my estimated and calculated values.
5 Magnetic Dipole
a) The last part of this project focused on particle motion due to a magnetic dipole. The
equation for the Earth’s dipole magnetic field was given as:
B =
µom
4πr3
2cos(θ)ˆr + sin(θ)ˆθ. (27)
Converting to Cartesian coordinates, I used the relations
ˆr = sin(θ)cos(φ)ˆx + sin(θ)cos(φ)ˆy + cos(θ)ˆz, (28)
8
and
ˆθ = cos(θ)cos(φ)ˆx + cos(θ)sin(φ)ˆy − sin(θ)ˆz, (29)
and
x = Rsin(θ)cos(φ), y = Rsin(θ)sin(φ), z = Rcos(θ). (30)
to solve for the magnetic dipole field. Plugging Equations 28-30 into Equation 27, I obtained
the expression
B =
µom
4π(x2 + y2 + z2)
5
2
[(3zx)ˆx + (3yz)ˆy + (2z2
− x2
− y2
)ˆz]. (31)
for the Earth’s magnetic dipole field in Cartesian coordinates.
b) In order to simulate the path of the particle along the Earth’s magnetic dipole, I set
the initial conditions to be at a distance of r = 2RE, the initial velocity components to be
vx0 = vy0 = vz0, the magnetic moment of the Earth to be M = 7.94e22 A m2
, and the total
kinetic energy to be 400 keV.
Figure 7: The path of the proton as it travels along the Earth’s dipole.
From Figure 7, I found the minima of the velocity to be around 7.00 s and 2.00 s which gave
me an estimated mirror bounce period of around (7.00 s) - (2.00 s) = 5.00 s. My estimated
mirror bounce period was 5.00 s.
c) I adjusted the run time in order to simulate the orbit of a particle around the Earth and
only plotted every thousandth point for simplicity.
9
Figure 8: The proton’s orbit around the Earth.
The simulation of the particle’s orbit around the Earth is shown in Figure 8. Looking at
where the point ended its orbit in the resulting plot, I came up with an estimated value for
period to be around 3400 s.
d) Computing the gradient-curvature drift, I started with the gradient-curvature equation
as provided by Hughes [2]:
vR + v B =
m
q
(v +
v2
⊥
2
)
Rc × B
R2
EB2
, (32)
where I still assumed r = Rc = 2RE and that the radius of curvature of the field line crossed
the equatorial plane at r = r0 = r0
3
. Plugging in the assumptions made and the parameters
used, the gradient-curvature drift resulted in the equation
vR + v B =
m
q
(v +
v2
⊥
2
)
2RE( µm
4πr3 )
. (33)
Using Equation 33, I would have obtained a value for the orbital period of the particle, but,
unfortunately, I could not figure out the calculation or obtain a reasonable answer. Given
the parameters and conditions, I imagine it would have been within a reasonable factor
compared to the estimated value.
References
[1] Bittencourt, J. A. ”Fundamentals of Plasma Physics — J. A. Bittencourt — Springer.”
J. A. Bittencourt — Springer. Springer-Verlag New York, n.d. Web.
10
[2] Hughes, J. M. ”Fundamentals of Space Physics”. Department of Physical Science. Embry-
Riddle Aeronautical University, n.d. Web.
[3] NOTE: A big thank you goes out to all of my fellow students who helped me with the
coding and allowed me to compare MATLAB data!
11

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Zontos___EP_410___Particle_Motion

  • 1. The Different Types of Particle Motion Zoe Zontos November 16, 2016 1 Introduction The purpose of this project was to investigate different types of particle motion. The study of the motion of particles in space physics is important and provides insight into understand- ing how dynamical processes in plasmas work. It also provides insight on the microscopic behavior of particles which is significant in order to understand plasma and space physics on a macroscopic level [1]. In this project, I worked with four types of particle motion: cyclotron motion, gradient-B drift motion, magnetic mirroring, and a magnetic dipole. 2 Cyclotron Motion a) I first investigated the cyclotron motion of a particle. The analytical velocity vector of a particle undergoing pure cyclotron was given as v(t) = (v⊥ sin(ωct))ˆx + (v⊥ cos(ωct))ˆy + (v )ˆz. (1) In order to determine the position of the particle as a function of time, I integrated v(t) to obtain t 0 v(t)dt = t 0 (v⊥ sin(ωct))ˆx + (v⊥ cos(ωct))ˆy + (v )ˆz)dt. (2) From this integration, the position vector with respect to time for the particle was determined to be r(t) = v⊥ ωc (1 − cos(ωct))ˆx + v⊥ ωc sin(ωct)ˆy + v tˆz. (3) Equation 3 represents the analytical solution for the position vector as a function of time for a particle undergoing pure cyclotron motion. (NOTE: The equation ωc = |q|B m represents the cyclotron frequency where q is the charge of the particle, B is the magnetic field value, and m is the mass of the particle [2].) b) I plotted the graph of the cyclotron motion of a proton showing the trajectories of the analytical solution along with the numerical solution as shown in Figure 1. For the proton, the mass is 1.672e-27 kg, the charge is 1.602e-19 C, and I used a magnetic field value of 45,000 nT. 1
  • 2. Figure 1: Analytical and numerical solutions for the motion and position of a proton through a uniform magnetic field. The resulting plot revealed that the particle trajectories for the analytical and numerical solutions were similar within a uniform magnetic field which was expected. c) Utilizing the same technique as in part b, I repeated the process and plotted the analytical and numerical solutions of cyclotron motion for an electron. For the electron, the mass is 9.109e-31 kg, the charge is the same as that of a proton, and the same magnetic field value was also used. 2
  • 3. Figure 2: Analytical and numerical solutions for the motion and position of a charged electron within a uniform magnetic field. The plot in Figure 2 revealed that, like the proton, the analytical and numerical solutions were similar for the electron within a uniform magnetic field as well. Figures 1 and 2 both showed that the direction of the proton and electron were clockwise and counterclockwise (respectively) due to the Lorentz force F = q(E + v × B) acting on the particles where E = 0 and a constant B field for both, but the charges have opposite signs. 3 Gradient-B Drift a) The next part of this project focused on the gradient-B drift motion of a proton. As explained in Hughes, the gradient-B drift has an E=0 and a perpendicular gradient present in the field strength. Wherever the magnetic field is changing, the particle will travel through as long as it has some initial perpendicular velocity [2]. Introducing a magnetic field with a gradient to the proton, I simulated and plotted the motion and position of the proton while under the influence of gradient-B drift motion alongside with its cyclotron motion. The equation for magnetic field was given as B(x, y, z) = Bo(1 + 1 L x)ˆz, (4) 3
  • 4. where the reference magnetic field used was Bo = 45,000 nT, the initial velocity in the y- direction was v(0) = 4000 m/s, and the L = 2 m. The magnetic field was kept strictly in the z-direction while also including a linear variation in order to maintain a constant gradient on the particle. Figure 3: Motion and position of a proton while under the influence of gradient-B drift motion and still undergoing cyclotron motion. The plot of the proton in Figure 3 shows that the trajectories of its cyclotron motion and gradient-B drift motion appear to be approximately equal. b) In order to determine the position vector of the particle, I started by using the provided equation for the magnetic field vector B = Bo(1 + 1 2 x)ˆz, (5) and the equation for the gradient-B drift which according to Hughes was given as v B = 1 2 v2 ⊥ ωc B × B B2 , (6) where B × B = − B2 o 2 (1 + 1 2 x)ˆy. (7) I plugged B × B into Equation 6 for v B which resulted in 4
  • 5. v B = mv2 ⊥B2 o (1 + 1 2 x) 4qB3 o (1 + 1 2 x)3 ˆy. (8) Simplifying the solution left the v B to be v B = mv2 ⊥ (1 + 1 2 x)3 ˆy. (9) I then integrated v B with respect to t and obtained an expression for the position r(t): r(t) = v⊥ ωc [1 − cos(ωct)]ˆx + [ v⊥ ωc sin(ωct) + mv2 ⊥ 4qBo(1 + 1 2 )x2 t]ˆy + v tˆz. (10) I plotted the analytical solution for r(t) which is seen in Figure 3 where the trajectories of the particles are relatively the same [2]. c) In order to demonstrate how changing the initial velocity affects the the gradient-B drift motion of the particle, I plotted the proton’s motion and position when its initial velocity was doubled (left) and halved (right) as shown in Figure 4. Figure 4: Motion and position of a proton while under the effects of gradient-B drift motion when its initial velocity is doubled (left) and halved (right). The results in Figure 4 revealed that changing the velocities of the particle affected the particle’s Larmour radius as well as its y-direction component of position. The y-direction of the particle’s position doubled when the velocity was doubled and halved when the particle’s velocity was halved. In the case where the initial velocity of the proton was doubled, the Larmour radius, rL = mv⊥ |q|B = v⊥ ωc , also doubled. Likewise, when the initial velocity was cut in half, the Larmour radius also decreased by half. 4 Magnetic Mirrors a, b) This part of the project focused on using a magnetic mirror to simulate a mirroring ’bottling’ motion undergone by the particle due to it moving through a region of a weak field 5
  • 6. surrounded by a strong field. Using Gauss’s Law, I determined the radial component of the magnetic field. Starting with the expanded form of Gauss’s Law in cylindrical coordinates: • B = 1 ρ ∂ ∂ρ (ρBρ) + ∂Bφ ρ∂φ + ∂Bz ∂z = 0, (11) I first solved for Bρ: ∂ ∂ρ (ρBρ) = −ρ ∂Bz ∂z , (12) which simplified to ρBρ = − ρ2 2 ∂Bz ∂z , (13) and even further Bρ = − ρ 2 ∂Bz ∂z . (14) I then rewrote Equation 14 as Bρ = − x2 + y2 Boz a2 . (15) Using the result found in Equation 14 and 15, I then converted to Cartesian coordinates which resulted in a Bx(x, y, z) and By(x, y, z) shown below. For Bx(x, y, z): Bx(x, y, z) = Bρ(z)cosθ, (16) which became Bx(x, y, z) = −x Boz a2 . (17) And for By(x, y, z): By(x, y, z) = Bρ(z)sinθ, (18) which became By(x, y, z) = −y Boz a2 . (19) The resulting magnetic field vector in Cartesian coordinates was B(x, y, z) = −x Boz a2 − y Boz a2 . (20) c) Choosing the origin as the initial position and using initial velocity components of vy0 = vz0= 10,000 m/s, I plotted the path of the particle to demonstrate that it mirrors correctly as shown in Figure 5. 6
  • 7. Figure 5: Mirroring of the particle with an initial velocity of 10,000 m/s and its initial position at the origin. The mirroring of the particle in Figure 5 was correct; the particle moved in a path that was ’mirrored’ on both sides (it moved symmetrically). Figure 6: The particle’s velocity versus time in the axis of the bottle (z-direction). 7
  • 8. The particle did mirror correctly, and I estimated the period of mirror motion by looking at the z-direction velocity of the particle inside the bottle in Figure 6. The bounce period minima appeared to be at around 0.280s and 0.075 s, so the estimated bounce period is (0.270 s) - (0.075 s) = 0.195 s which resulted in an estimated frequency of (1/0.195 s) = 5.13 Hz. d) Using the equation for the mirror force, I solved for the the parallel equation of motion for the specific magnetic field used. The mirror force provided was written as Fz = −µ ∂Bz ∂z = m d2 z dt2 , (21) where the magnetic moment µ = 1 2 mv2 ⊥ B , so, taking the second derivative, I determined d2 z dt2 = − mv2 ⊥ mBo ( 2Boz a2 ) = −v2 ⊥ z a2 . (22) Solving for the differential equation yields: z(t) = Asin( v⊥t a ) + Bcos( v⊥t a ). (23) I applied the initial conditions of z(o) and z(t) to find that z(0) = Bcos( v⊥t a ), (24) and z(t) = Asin( v⊥t a ). (25) The resulting equation for the angular frequency can be written as f = v⊥ 2πa . (26) Solving for the frequency and period, I obtained a value of 5.305 Hz for the frequency and a period of t = 0.189 s. Comparing the calculated values to that of the estimated values, there was a 3.17-3.29 % difference between my estimated and calculated values. 5 Magnetic Dipole a) The last part of this project focused on particle motion due to a magnetic dipole. The equation for the Earth’s dipole magnetic field was given as: B = µom 4πr3 2cos(θ)ˆr + sin(θ)ˆθ. (27) Converting to Cartesian coordinates, I used the relations ˆr = sin(θ)cos(φ)ˆx + sin(θ)cos(φ)ˆy + cos(θ)ˆz, (28) 8
  • 9. and ˆθ = cos(θ)cos(φ)ˆx + cos(θ)sin(φ)ˆy − sin(θ)ˆz, (29) and x = Rsin(θ)cos(φ), y = Rsin(θ)sin(φ), z = Rcos(θ). (30) to solve for the magnetic dipole field. Plugging Equations 28-30 into Equation 27, I obtained the expression B = µom 4π(x2 + y2 + z2) 5 2 [(3zx)ˆx + (3yz)ˆy + (2z2 − x2 − y2 )ˆz]. (31) for the Earth’s magnetic dipole field in Cartesian coordinates. b) In order to simulate the path of the particle along the Earth’s magnetic dipole, I set the initial conditions to be at a distance of r = 2RE, the initial velocity components to be vx0 = vy0 = vz0, the magnetic moment of the Earth to be M = 7.94e22 A m2 , and the total kinetic energy to be 400 keV. Figure 7: The path of the proton as it travels along the Earth’s dipole. From Figure 7, I found the minima of the velocity to be around 7.00 s and 2.00 s which gave me an estimated mirror bounce period of around (7.00 s) - (2.00 s) = 5.00 s. My estimated mirror bounce period was 5.00 s. c) I adjusted the run time in order to simulate the orbit of a particle around the Earth and only plotted every thousandth point for simplicity. 9
  • 10. Figure 8: The proton’s orbit around the Earth. The simulation of the particle’s orbit around the Earth is shown in Figure 8. Looking at where the point ended its orbit in the resulting plot, I came up with an estimated value for period to be around 3400 s. d) Computing the gradient-curvature drift, I started with the gradient-curvature equation as provided by Hughes [2]: vR + v B = m q (v + v2 ⊥ 2 ) Rc × B R2 EB2 , (32) where I still assumed r = Rc = 2RE and that the radius of curvature of the field line crossed the equatorial plane at r = r0 = r0 3 . Plugging in the assumptions made and the parameters used, the gradient-curvature drift resulted in the equation vR + v B = m q (v + v2 ⊥ 2 ) 2RE( µm 4πr3 ) . (33) Using Equation 33, I would have obtained a value for the orbital period of the particle, but, unfortunately, I could not figure out the calculation or obtain a reasonable answer. Given the parameters and conditions, I imagine it would have been within a reasonable factor compared to the estimated value. References [1] Bittencourt, J. A. ”Fundamentals of Plasma Physics — J. A. Bittencourt — Springer.” J. A. Bittencourt — Springer. Springer-Verlag New York, n.d. Web. 10
  • 11. [2] Hughes, J. M. ”Fundamentals of Space Physics”. Department of Physical Science. Embry- Riddle Aeronautical University, n.d. Web. [3] NOTE: A big thank you goes out to all of my fellow students who helped me with the coding and allowed me to compare MATLAB data! 11