Physics 430: Lecture 5 Complex Exponentials and Rockets Dale E. Gary NJIT Physics Department

September 15, 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 where q is the charge and B is the magnetic field strength. The equation of motion then becomes 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: and the velocity can in general have any direction. Hence, and the three components of the equation of motion are:

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, v z = 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 (v x, v y ) = transverse velocity.  To simplify, we define the parameter  = qB/m :, so the equations of motion become:  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 i is the square root of  1. Let us define: and then plot the value of  as a vector in the complex plane whose components are v x and v y. imaginary part real part vxvx vyvy  = v x + iv y

September 15, 2009 Motion of a Charge in a Uniform Magnetic Field-3  Next, we take the time derivative of  : or  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  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  We introduced the idea of using a complex number made of the components of velocity, v x and v y, to solve the coupled pair of equations  i.e. we write, which can be represented as a vector in the complex plane:  We saw that with this substitution the equations combine into a single equation, with solution:.  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). 2.6 Complex Exponentials imaginary part real part vxvx vyvy  = v x + iv y

September 15, 2009  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  Notice that this agrees with the expectation that it equals its own derivative. You can also show that which means that Ae kz is the solution to.  This is the same form as, so long as we identify k =  i .  Obviously, the expansion of Taylor Series Expansion of e z

September 15, 2009 Taylor Series Expansion of e z  Let’s look at this rearrangement a bit closer:  So we are left with the identity:  Note that, when plotted as a vector in the complex plane, this is a unit- length vector. Any complex number can thus be written in terms of this: Taylor series expansion for cos  Taylor series expansion for sin  Even powers Odd powers Try problem 2.18 for yourself

September 15, 2009 Graphical Representation of e i   Plotting Euler’s formula in the complex plane  What is the graphical representation of ? Note that in general, the constant A may itself be complex. If so, it can be written in the form where by definition the amplitude is and the “phase” is  Combining these, we have whose graph is above  cos  sin  x y eiei 1  tt A = ae i  x y a  =Ae  i  t

September 15, 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, which yields the equations of motion for the three components [with B = (0, 0, B) ]:  For the x and y components, this gave, with  = qB/m. We also defined, and found that with this definition the problem reduced to a single equation, whose solution is.  Using the complex variable ideas that we just introduced, we see that this last expression for  can be represented as a rotating vector in the complex plane, rotating at angular speed , whose components are the velocities v x and v y.  = v x + iv y

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, once, which gives:  But what is  ? It is another complex quantity, whose components are the x and y positions,.  Let us write the combination of constants as a new constant C, so that  What is the meaning of the integration constant that we wrote simply as “constant?” As usual, it is related to our choice of coordinates, and it is itself complex. However, we can choose it to be zero.  = x + iy x y  tt C = ce i  c  =Ce  i  t  =Ce  i  t + constant X+iY

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:  Note, however, that the position at t = 0 is not zero, but rather is  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 x o = y o = 0 ? Well, if we did set x o = y o = 0, then the particle would be at the origin at t = 0.  So, setting our coordinate system at the center of the circle is not the same as setting it to the position of the particle at t = 0.  To finish this discussion, recall that the z -component of velocity was constant, so the particle executes a helical motion, like a corkscrew, along the direction of B. x y  tt c

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.  The circular motion (or gyration) occurs at the angular speed  = qB/m, and is called the gyrofrequency (also called the cyclotron frequency, named for the device called the cyclotron).  The radius of the gyration (called the gyroradius) is r = v/  = mv/qB.  Generally, the charged particle is constrained to stay close to a B field line. This has great consequences for plasmas like the solar corona. x y z x-y-z (helical) path B

September 15, 2009  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:  This is expressed by the relation, where is the total momentum of the system and F ext 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. 3.1 Conservation of Momentum If the net external force F ext on an N-particle system is zero, the system’s total momentum P =  m  v  is constant. Principle of Conservation of Momentum

September 15, 2009  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? 3.2 Rockets  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.  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  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? How Do Rockets Work?  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 by Newton’s third law (action and reaction), or equivalently, by the law of conservation of momentum.  The rocket works by throwing mass in the backward direction, in order to give it forward momentum. The key, of course, is to throw the mass backward with the maximum possible speed (thereby maximizing the momentum)—hence the high-power rocket motor. There is a major complication, though—the mass of the rocket changes as the fuel is spent.

September 15, 2009  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 P rocket (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  v ex relative to the ground, where v ex is the exhaust velocity relative to the rocket. The exhaust momentum, then, is P exhaust (t + dt) =  dm(v  v ex ).  Adding these, the total momentum at time t + dt is: P(t + dt) = (m + dm)(v + dv)  dm(v  v ex ) = mv + m dv + dm v ex, 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 v ex The Rocket Problem-1

September 15, 2009  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 v ex or, dividing through by dt (recall my caution about doing this), we have the equation of motion  The force is called the thrust. Recall that we defined as negative, so thrust is positive. This should make perfect sense—the rocket is losing mass.  To solve this equation, we write it in separable form where now upon integration the mass dependence is logarithmic: Here, v o is the initial velocity and m o the initial mass of the rocket. The Rocket Problem-2

September 15, 2009  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  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 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 0.6m o.) 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 v ex. (b) Suppose instead it burns the fuel in two stages as follows: In the 1 st stage it burns a mass 0.3m o of fuel. It then jettisons the 0.1m o mass 1 st stage fuel tank and then burns the remaining 0.3m o of fuel. Find the final speed in this case, assuming the same value of v ex throughout, and compare. The Rocket Problem-3

September 15, 2009  Part (a) of the problem is trivial. The final speed is:  For part (b), we find the speed at the time of the first stage burnout: and use this as the initial speed for stage 2.  After jettisoning the 0.1m o spent rocket, the second stage initial mass (rocket plus fuel) is 0.6m o, and the final mass is 0.3m o. At the time of second stage burnout, then, we have a final speed  Thus, there is a considerable speed boost from the 2-stage rocket, for the same expenditure of fuel. Problem 3.12—Solution