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Physics 430: Lecture 5
Complex Exponentials and
Rockets
Dale E. Gary
NJIT Physics Department
2.5 Motion of a Charge in a
Uniform Magnetic Field

You may recall from Physics 121 that the force on a charge moving in a
magnetic field is
F  qv  B
where q is the charge and B is the magnetic field strength. The equation of
motion then becomes
mv  qv  B
which is a first-order differential equation in v.
 In this type of problem, we are often free to choose our coordinate system
so that the magnetic field is along one axis, say the z-axis:
B  (0,0, B )
and the velocity can in general have any direction v  (vx , v y , vz ). Hence,
v  B  (v y B,  v x B, 0)
and the three components of the equation of motion are: mvx  qv y B
mv y  qvx B
mvz  0
September 15, 2009
Motion of a Charge in a Uniform
Magnetic Field-2
This last equation simply says that the component of velocity along B,
vz = const. Let’s now focus on the other two components, and ignore the
motion along B. We can then consider the velocity as a two-dimensional
vector (vx, vy) = transverse velocity.
 To simplify, we define the parameter w = qB/m:, so the equations of motion
become:
vx  w v y



v y  w v x
We will take the opportunity provided by these two coupled equations to
introduce a solution based on complex numbers.
As you should know, a complex number is a number like z = x + iy, where
imaginary
i is the square root of 1. Let us define:
part
h = vx + ivy
and then plot the value of h as a vector in the
vy
complex plane whose components are vx and vy.
real part
vx
September 15, 2009
Motion of a Charge in a Uniform
Magnetic Field-3

Next, we take the time derivative of h:
h  vx  iv y  wv y  iwvx  iw (vx  iv y )
or
h  iwh
So the equation in terms of this new relation has the same form we saw in
the previous lecture for linear air resistance, with the familiar solution
h  Aeiwt
 The only difference is that this time the argument of the exponential is
imaginary, but it turns out that this makes a huge difference.
 Before we can discuss the solution in detail, however, we need to introduce
some properties of complex exponentials.

September 15, 2009
2.6 Complex Exponentials


We introduced the idea of using a complex number made of the
components of velocity, vx and vy, to solve the coupled pair of equations
vx  w v y
v y  w v x
i.e. we write h = vx + ivy , which can be represented as a vector in the
complex plane:
imaginary
part
vy
real part


vx
We saw that with this substitution the equations combine into a single
equation h  iwh, with solution: h  Aeiwt .
We now want to take a closer look at complex exponentials, which have
wide applicability in a field of study called complex variables (e.g. Math
332).
September 15, 2009
Taylor Series Expansion of ez
We have seen that any function can be expanded in terms of its derivatives
in the Taylor Series, which in the case of the exponential function gives an
expansion in terms of powers
1
1
ez  1 z  z2  z3  
2!
3!
 Notice that this agrees with the expectation that it equals its own
derivative. You can also show that
d
Aekz  k Aekz
dt
d
which means that Aekz is the solution to
f  kf or f  kf .
dt


 

This is the same form as h  iwh, so long as we identify k = iw.
1
1
1
2
3
4
 Obviously, the expansion of
e i  1  i  i   i   i   
2!
3!
4!
 2 4
  3

 1 

   i    
3!
 2! 4!
 


September 15, 2009
Taylor Series Expansion of ez

Let’s look at this rearrangement a bit closer:
 2 4
  3

e  1      i    
3!
 2! 4!
 

i
Even powers
Odd powers
Taylor series expansion
for cos 
Taylor series expansion
for sin 
Try problem 2.18
for yourself
So we are left with the identity:
ei  cos  i sin  [Euler' s formula]
 Note that, when plotted as a vector in the complex plane, this is a unitlength vector. Any complex number can thus be written in terms of this:

Aei  A cos   iA sin 
September 15, 2009
Graphical Representation of ei

Plotting Euler’s formula ei  cos   i sin  in the complex plane
y
e i
sin 
1
cos 
A = aeid
a
x

y
d
h=Aeiwt
x
wt
What is the graphical representation of h  Aeiwt ? Note that in general, the
constant A may itself be complex. If so, it can be written in the form
A  Areal iAimag  a cos d  ia sin d  aeid
2
2
 Aimag
where by definition the amplitude is a  Areal
and the “phase” is
A
d  arctan imag
Areal
 iwt
 aeid e iwt  aei (d wt ) whose graph is above
 Combining these, we have h  Ae

September 15, 2009
2.7 Solution for Charge in B Field

We are now ready to return to the problem, to look at the motion of a charge
in a B field. To quickly remind you, we started with the magnetic force on the
charge F  qv  B , which yields the equations of motion for the three
components [with B = (0, 0, B)]:
mvx  qv y B
mv y  qvx B

mvz  0
v  w v y
For the x and y components, this gave x
, with w = qB/m. We also
v y  w v x
defined h = vx + ivy , and found that with this definition the problem reduced to
a single equation h  iwh , whose solution is h  Aeiwt.
 Using the complex variable ideas that we just introduced, we see that this last
expression for h can be represented as a rotating vector in the complex plane,
rotating at angular speed w, whose components are the velocities vx and vy.
September 15, 2009
Trajectory of the Charge-1

To get a complete solution, we need to know the changing position vector r(t).
It should be obvious that we need to integrate h  Aeiw,t once, which gives:
x  hdt   Aeiwt dt




iA
w
e iwt  constant
But what is x? It is another complex quantity, whose components are the x and
y positions, x = x + iy .
iA
Let us write the combination of constants
as a new constant C, so that
w
y
 iwt
x  Ce  constant
C = ceid
c
X+iY
What is the meaning of the integration constant
x
d
that we wrote simply as “constant?” As usual, it
x=Ceiwt wt
is related to our choice of coordinates, and it is
x=Ceiwt + constant
itself complex. However, we can choose it to be zero.
September 15, 2009
Trajectory of the Charge-2





So we see that the (x, y) part of the trajectory is a circle that may be offset
from our coordinate system. However, choosing our coordinate system origin
to be the center of the circle, we simply have:
x  x  iy  Ceiwt
Note, however, that the position at t = 0 is not zero, but rather is
x (0)  xo  iyo  C
This may seem confusing at first. If we chose our coordinate system to be the
center of the circle, why doesn’t this set the initial conditions to be xo = yo = 0?
Well, if we did set xo = yo = 0, then the particle would be at the origin at t = 0.
So, setting our coordinate system at the center of the
y
circle is not the same as setting it to the position of
c
the particle at t = 0.
x
d
To finish this discussion, recall that the z-component
wt
of velocity was constant, so the particle executes a
helical motion, like a corkscrew, along the direction of B.
September 15, 2009
Trajectory of the Charge-3
Altogether, the path looks something like the drawing below. The motion is
circular in the x-y plane, and uniform in the z direction, along the magnetic
field.
B
 The circular motion (or gyration) occurs at the
angular speed w = qB/m, and is called the
x-y-z (helical) path
gyrofrequency (also called the cyclotron frequency,
named for the device called the cyclotron).
y
z
 The radius of the gyration (called the gyroradius)
is r = v/w = mv/qB.
x
 Generally, the charged particle
is constrained to stay close to
a B field line. This has great
consequences for plasmas like
the solar corona.

September 15, 2009
3.1 Conservation of Momentum

We are now switching gears completely, on to Chapter 3. This first section
returns to our discussion of conservation of momentum from the first
lecture:
Principle of Conservation of Momentum
If the net external force Fext on an N-particle system is zero,
the system’s total momentum P = Smava is constant.
This is expressed by the relation P  Fext , where P  p1  p2    pn   pa
is the total momentum of the system and Fext is the sum of all external
forces.
 Such conservation principles are very powerful, and we can learn a lot
about the behavior of systems of particles solely from them.
 In lecture 1 we did Problem 1.30, the inelastic collision of two masses (see
the end of lecture 1). Example 3.1 of the text shows a slightly more
complicated version of the same problem (where the second mass does not
start at rest). The approach is to set the momentum before the collision
equal to the momentum after the collision.

September 15, 2009
3.2 Rockets
We would now like to look at a somewhat more complicated, but
interesting, application of the conservation of momentum—that of a rocket.
 To introduce this subject, imagine that you are floating in space (hopefully
in a spacesuit), and you find yourself motionless with respect to your space
station’s airlock, say 100 m away. In fact, your hand-held laser meter tells
you that the distance is 97.3 m, exactly. What can you do to get yourself
back to the airlock?

What you want to do is throw your laser meter in the direction opposite the
airlock. By the conservation of momentum, the laser meter imparts a
momentum to you equal and opposite to the momentum you give to the
laser meter. Another way to say it is that the total momentum of you and
the laser meter is zero (relative to the air lock) before you make the throw,
and it remains zero afterward.
mlm
v


v lm
you
0  myou v you  mlm v lm
myou
 What happens if the laser meter slips as you throw it, and, although it has
the right direction, it is only moving very slowly?

September 15, 2009
How Do Rockets Work?

When a rocket is launched, it appears that the rocket
exhaust is pushing against the ground, making the rocket go
up. What is wrong with this idea?
For this explanation to work, once the rocket got away from
the ground it would cease to work. One could always argue
that it can still push against the air. But what about when it
gets into space? There is nothing to push against, yet the
rocket still works.
 The answer, as you no doubt know, is that the rocket works
There is a major
by Newton’s third law (action and reaction), or equivalently,
complication,
by the law of conservation of momentum.
though—the mass
 The rocket works by throwing mass in the backward
of the rocket
direction, in order to give it forward momentum. The key, of
changes as the fuel
course, is to throw the mass backward with the maximum
is spent.
possible speed (thereby maximizing the momentum)—hence
the high-power rocket motor.

September 15, 2009
The Rocket Problem-1
We would like to discuss the motion of rockets, which involves the
conservation of momentum. Let’s work in a reference frame (coordinate
system) fixed to a point on the ground. Consider that at time t, the
momentum of the rocket plus unspent fuel is P(t) = mv. A short time dt later,
the momentum is now Procket(t + dt) = (m + dm)(v + dv), where obviously dm is
negative.
 The fuel ejected during this time dt has mass dm (a quantity that is positive),
and velocity v  vex relative to the ground, where vex is the exhaust velocity
relative to the rocket. The exhaust momentum, then, is
Pexhaust(t + dt) = dm(v  vex).
 Adding these, the total momentum at time t + dt is:
P(t + dt) = (m + dm)(v + dv)  dm(v  vex) = mv + m dv + dm vex,
where I drop the doubly small product dm dv. This is an extremely common
thing to do, and is called linearizing the problem. If we do not do this, the
problem becomes non-linear and generally must be solved numerically.
 The change in momentum is then P(t + dt)  P(t) = m dv + dm vex

September 15, 2009
The Rocket Problem-2
Applying conservation of momentum, this change in momentum must be
zero. But remember, there is a condition under which we are allowed to
employ conservation of momentum. It only holds when all external forces are
zero. We will use it here, but it amounts to ignoring gravity, which clearly is a
present external force. You can do the problem with gravity (Problem 3.11).
 Ignoring gravity, then, we have m dv = dm vex or, dividing through by dt (recall
my caution about doing this), we have the equation of motion
 vex
mv  m
 vex is called the thrust. Recall that we defined m
 as negative, so
 The force  m
thrust is positive. This should make perfect sense—the rocket is losing mass.
dm
 To solve this equation, we write it in separable form
dv  vex
m

where now upon integration the mass dependence is logarithmic:
v  vo  vex ln( mo / m)
Here, vo is the initial velocity and mo the initial mass of the rocket.
September 15, 2009
The Rocket Problem-3
The interesting thing about this result is that by far most of the mass has to be
thrown away in order for a rocket to work. Say the rocket fuel is 90% of the
total initial mass (leaving only 10% for the rocket itself and the payload).
 Then the absolute maximum speed of such a rocket is only v  vex ln( 10)  2.3 vex
 You can see that having a high exhaust speed is essential, but even then,
another trick is needed to reach high velocities. The trick is to use multistage
rockets. Dropping a spent first stage reduces the initial and final masses of the
stage 2 rocket by the same amount.
 As an illustration, let’s do Problem 3.12. Here is the statement of the problem:


To illustrate the use of a multistage rocket consider the following: (a) A certain rocket
carries 60% of its initial mass as fuel. (That is, the mass of fuel is 0.6mo.) What is the
rocket’s final speed, accelerating from rest in free space, if it burns all its fuel in a
single stage? Express your answer as a multiple of vex. (b) Suppose instead it burns
the fuel in two stages as follows: In the 1st stage it burns a mass 0.3mo of fuel. It
then jettisons the 0.1mo mass 1st stage fuel tank and then burns the remaining 0.3mo
of fuel. Find the final speed in this case, assuming the same value of vex throughout,
and compare.
September 15, 2009
Problem 3.12—Solution
Part (a) of the problem is trivial. The final speed is:
v  vex ln( mo / m)  vex ln( 1 / 0.4)  0.92 vex
 For part (b), we find the speed at the time of the first stage burnout:


v1st  vex ln( mo / m)  vex ln( 1 / 0.7)  0.36 vex
and use this as the initial speed for stage 2.
After jettisoning the 0.1mo spent rocket, the second stage initial mass (rocket
plus fuel) is 0.6mo, and the final mass is 0.3mo. At the time of second stage
burnout, then, we have a final speed
v  v1st  vex ln( 0.6mo / 0.3mo )
 v1st  vex ln( 2)  0.36 vex  0.69 vex

 1.05 vex
Thus, there is a considerable speed boost from the 2-stage rocket, for the same
expenditure of fuel.
September 15, 2009