Section 10–3 Momentum is conserved

(Center of mass frame / Laboratory frame / Moving car’s frame)

In this section, Feynman discusses collisions by comparing “view from moving car” with “view from center of mass,” “view from laboratory,” and “view from CM system”. Essentially, the discussions include three reference frames: center of mass frame, laboratory frame, and moving car’s frame. The title of this section could be rephrased as “three reference frames of collisions”.

1. Center of mass frame:

“Two views of an inelastic collision between equal masses (Feynman et al., 1963, section 10–3 Momentum is conserved).”

Feynman mentions the need for an assumption that physical laws remain the same whether we are standing still or observing in a moving car. This assumption is applicable to collisions such as two moving objects collide and stick together. The collision in the example is also physically equivalent to an object, initially moving with velocity 2v, hits a stationary object that has the same mass. Interestingly, Feynman shows how the velocities of the two objects (after a collision) can be deduced without using the concept of linear momentum. In essence, he uses simple logic and two different frames of reference: center of mass frame and a moving car’s frame.

In Fig. 10.4, the phrase “view from center of mass” is used to describe a perspective from the center of mass frame. This frame of reference refers to an inertial frame in which the center of mass is fixed at the origin. Center of mass frame is also known as “zero momentum frame” in the sense that the total linear momentum of a system of particles is zero (Williams, 2002). An advantage of zero momentum frame is the convenience of viewing the collision such that it simplifies the calculations. Mathematically, if the first particle approaches the collision point (or the second particle) with linear momentum p, then the second particle will approach the same point with linear momentum -p.

2. Laboratory frame:

“When it is all finished they will be moving at ½(v₁−v₂) with respect to the car. What then is the actual speed on the ground? (Feynman et al., 1963, section 10–3 Momentum is conserved).”

In another inelastic collision, we assume that two identical objects initially move with velocities v₁ and v₂ before they stick together. If we observe the collision in a car, say moving at velocity v₂, one of the objects would appear to be at rest and the other would appear to have a velocity of v₁−v₂. After the collision, they are expected to be moving at an average velocity of ½(v₁−v₂) with respect to the moving car or ½(v₁−v₂) + v₂ with respect to the ground. Feynman uses Galilean principle of relativity again to deduce the final velocity of the two objects without first using a mathematical formula involving linear momentum. Note that he uses the laboratory frame and moving inertial frame instead of the center-of-mass frame in this example.

In Fig. 10.5, the phrase “view from lab” is used to describe the laboratory frame of reference. In short, we may use the term laboratory frame or Earth’s frame to describe a frame of reference that views from a laboratory on the Earth in which an experiment is performed. For reasons of simplicity, physicists may assume the laboratory frame to be an inertial frame that is at rest (French, 1971; Morin, 2008). In particle physics, one may consider the laboratory frame in which the detectors for a particle accelerator are stationary with respect to the Earth. To be more precise, physicists may state that the laboratory is rotating with the Earth about its axis. Thus, a laboratory frame can also be defined as a non-inertial frame or a rotating frame of reference (Ryder, 2008).

3. Moving car’s frame:

“This is very easy to answer using our principle of Galilean relativity, for we simply watch the collision which we have just described from a car moving with velocity −v/2 (Feynman et al., 1963, section 10–3 Momentum is conserved).”

Feynman discusses another inelastic collision problem in which the two objects have different masses. The question is what will happen if an object that has mass m moves with velocity v hits a stationary object that has mass 2m? Based on Feynman’s thought experiment, the masses m and 2m attain velocities –v and v/2 respectively. He uses “view from CM system” (instead of “view from center of mass”) first followed by “view from car.” In other words, it is insightful to visualize and solve the collision problem by using the “zero momentum frame” and “moving car’s frame.” Perhaps Feynman could be more explicit and explain why he adopts the two different reference frames.

In Fig. 10.7, the phrase “view from car” is used to describe an inertial frame of reference based on an observer that is in a car moving at constant velocity. We may use the term “moving car’s frame” to describe an inertial frame in which an observer views one of the objects to be stationary (as a target). Thus, the momentum of an object becomes zero and it slightly simplifies the calculation needed. Interestingly, a definition of an inertial frame is based on the validity of Newton’s first law (Kleppner & Kolenkow, 2014). Philosophers may question this definition because Newton’s laws are also defined in terms of the inertial frame. More importantly, Feynman has explained that the conservation of momentum is supported by an experiment using an air trough (instead of dependent on how we define the inertial frame).

Questions for discussion:

1. What is/are the advantage(s) of using a center of mass frame of reference (or zero momentum frame)?

2. What is/are the advantage(s) of using a laboratory frame of reference (or Earth’s frame)?

3. What is/are the advantage(s) of using a moving car’s frame of reference?

The moral of the lesson: it is possible to deduce the speed of an object after a collision by using three different frames of reference and without involving calculations based on the law of conservation of momentum.

References:

1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

2. French, A. (1971). Newtonian Mechanics. New York: W. W. Norton.

3. Kleppner, D., & Kolenkow, R. (2014). An Introduction to Mechanics (2nd ed.). Cambridge: Cambridge University Press.

4. Morin, D. (2008). Introduction to Classical Mechanics. Cambridge: Cambridge University Press.

5. Ryder, P. (2007). Classical Mechanics. Aachen: Shaker Verlag.

6. Williams, W. S. C. (2002). Introducing Special Relativity. London: Taylor & Francis.

Section 10–2 Conservation of momentum

(External forces / Point of view / Definition of equal masses)

In this section, the three interesting points are external forces, point of view, and definition of equal masses. Alternatively, this section could be revised and titled as “three conditions of conservation of linear momentum”: (1) no external forces; (2) frame of reference; and (3) conservation of mass.

1. External forces:

“…if there are no forces from the outside (external forces), there are no forces that can change the total momentum; hence the total momentum is a constant (Feynman et al., 1963, section 10–2 Conservation of momentum).”

Mathematically, Feynman shows that the total linear momentum of two particles remains constant even though there are mutual interactions between them. Generally speaking, all the internal forces (due to mutual interactions) between many particles will cancel out each other such that they do not change the total linear momentum of the particles. In essence, there is a law of linear momentum: the total linear momentum of a system of particles remains unchanged provided there is an absence of external forces. However, one may expect Feynman to elaborate that this law is an idealization because there is no isolated system of particles in the real world. For practical purposes, we assume gravitational forces and electromagnetic forces acting on the particles to be negligible.

In a sense, it is about an idealized condition necessary for the conservation of linear momentum of a system: the net force on the particles is zero. The concept of net force may be labeled as an external force or internal force depending on how we define a system of particle(s). If we focus on a system of two particles, we may consider the opposite forces acting on the two particles as internal forces that are equal and opposite in direction. In this case, the linear momentum of the system of two particles remains conserved. Alternatively, one may treat the two particles separately as two (one-particle) systems and label the opposite forces as external forces that can accelerate the two particles in opposite directions. In other words, the momenta of the two systems can be changed as a result of the so-called external forces.

2. Point of view:

“…we shall discuss the laws of impacts and collisions from a completely different point of view (Feynman et al., 1963, section 10–2 Conservation of momentum).”

Feynman explains that he will discuss the laws of impacts and collisions from a completely different point of view. It is based on Galilean principle (or principle of relativity) in which the laws of physics remain unchanged whether we are stationary or moving with a uniform speed in a straight line. Feynman illustrates this principle by giving an example where a child bouncing a ball in an airplane finds that the ball bounces in the same way as if it is bouncing on the ground. However, one may use the term inertial frame of reference instead of a point of view. That is, we are not simply viewing the bouncing of a ball in different directions (e.g., x-axis or y-axis).

An inertial frame of reference is also an idealized condition such that the linear momentum of a system of particles is conserved. This condition is idealized in the sense that there is no external force acting on any particles in an inertial reference frame. In the words of French (1971), “[t]here exist certain frames of reference with respect to which the motion of an object, free of all external forces, is a motion in a straight line at constant velocity (p. 162).” To be more accurate, there are external forces such as gravitational forces in a laboratory frame of reference (or simply Earth’s frame). The choice of a particular reference frame is usually for the purpose of problem-solving or a matter of convenience.

3. Definition of equal masses:

“…there are some physical laws involved, and if we accept this definition of equal masses, we immediately find one of the laws, as follows (Feynman et al., 1963, section 10–2 Conservation of momentum).”

Feynman suggests that we can use collision experiments to define equal masses. Firstly, two objects made of different materials, say copper and aluminum, may acquire equal (separation) speeds in a collision experiment if they have the same mass. Additionally, if one of two objects has the same masses as a third object (by measuring the velocities based on the same experiment), then we may infer that they all have the same mass. As a suggestion, one may explain that this is a thought experiment that cannot be performed easily and exactly. Interestingly, Feynman explained earlier that an object may gain mass when it moves at a faster speed. More important, the object may gain internal energy (e.g., the molecules in a ball move faster) through a collision or explosion such that its mass is increased.

Alternatively, we can suggest another idealized condition for the law of conservation of momentum: conservation of mass. For example, Sartori (1996) states that conservation of mass is a necessary condition in order for conservation of momentum to be covariant in the context of Galilean relativity. Furthermore, in Wilczek’s (2004) words, “the energies and momenta of such particles are given in terms of their masses and velocities, by well-known formulas, and we constrain the motion by imposing conservation of energy and momentum. In general, it is simply not true that the sum of the masses of what goes in is the same as the sum of the masses of what goes out (p. 10).” That is, we assume the masses of objects to be constant such that we can apply the equation, m₁v₁ + m₂v₂ = constant, in simple collision situations.

Questions for discussion:

1. Why is the absence of external forces needed as an idealized condition of the law of conservation of linear momentum?

2. Would you include an inertial frame of reference (or point of view) as another idealized condition of the law of conservation of linear momentum?

3. Would you consider conservation of mass as an important condition of the law of conservation of linear momentum?

The moral of the lesson: the law of conservation of linear momentum is dependent on the validity of idealization conditions.

References:

1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

2. French, A. (1971). Newtonian Mechanics. New York: W. W. Norton.

3. Sartori, L. (1996). Understanding relativity: a simplified approach to Einstein's theories. California: University of California Press.

4. Wilczek, F. (2004). Whence the Force of F = ma? II: Rationalizations. Physics Today, 57(12), 10-11.

Section 10–1 Newton’s Third Law

(Mathematical analysis / Three-body problem / Action equals reaction)

In this section, the three interesting points discussed are mathematical analysis, three-body problem, and action equals reaction.

1. Mathematical analysis:

“First, there are quite simple cases of motion which can be analyzed not only by numerical methods, but also by direct mathematical analysis (Feynman et al., 1963, section 10–1 Newton’s Third Law).”

It seems that Feynman tries to fill a gap of the previous lecture although he is supposed to introduce Newton’s third law of motion. Firstly, he mentions that there are simple cases of motion which can be analyzed not only by numerical methods, but also by mathematical analysis. As an example, it is possible to use mathematical analysis to show a planet’s orbit is elliptical instead of using numerical methods as discussed towards the end of the previous lecture. Feynman is referring to the use of a central force that is inversely proportional to the square of the distance (between the Sun and a planet). This two-body problem is sometimes known as the Kepler problem that can be solved by integration using polar coordinates (r and θ).

Instead of discussing Newton’s second law of motion in the previous lecture, there could be some discussions on how Newton developed his third law. In one of his scholium of the Principia, he acknowledges the contributions of Sir Christopher Wren, Dr. Wallis, and Huygen. In his own words, Newton (1687) writes that “Sir Christopher Wren confirmed the truth of the thing before the Royal Society by the experiment of pendulums, which Mr. Mariotte soon after thought fit to explain in a treatise entirely upon that subject (p. 26).” In addition, Newton discusses the experiment related to collisions of hanging pendulums.

2. Three-body problem:

“That is the famous three-body problem, which so long challenged human powers of analysis… (Feynman et al., 1963, section 10–1 Newton’s Third Law).”

Feynman possibly fills another gap of the previous lecture by discussing the three-body problem, which challenged human powers of analysis for a long time. For example, if there are two bodies going around the sun, there is no simple formula for the motions of the three bodies. Furthermore, there are situations such as the motion of the stars in a globular cluster where both methods fail. Although we can use direct mathematical analysis or numerical methods to solve simpler problems, it is still challenging (if not impossible) to achieve exact solutions for very complicated problems involving many bodies. Interestingly, Feynman bridges the difficulties of solving Newton’s second law to the need of principle of conservation of linear momentum that is the foundation of Newton’s third law.

The three-body problem can be cited as a classic example of a chaotic system. This problem dates back to the 1680s when Newton shows that his law of gravitation could predict the orbit of two bodies held together by gravity—such as the Sun and a planet. In 1889, King Oscar II of Sweden offered a prize to anyone who could solve the three-body problem. Henri Poincaré won the King’s prize without solving the general problem. Poincaré simplified the problem by considering two enormous bodies and a third that was very small: the small particles would move in a complicated (or chaotic) manner around the two large astronomical bodies. Recently, Šuvakov and Dmitrašinović (2013) identify 13 new families of solutions by tweaking an existing solution on a computer simulation.

3. Action equals reaction:

“This principle is that action equals reaction (Feynman et al., 1963, section 10–1 Newton’s Third Law).”

According to Feynman, the principle of Newton’s third law is mainly action equals reaction. In other words, if particle A exerts a force (F_BA) on particle B, then particle B will push on particle A with an equal force (F_AB) and in the opposite direction. This pair of forces effectively acts in the same line. Alternatively, one may explain that there are five features of Newton’s third law: (1) forces occur in pairs; (2) the nature of the two forces are the same (e.g., gravitational or electromagnetic); (3) both forces have the same magnitude; (4) both forces acts along the same line but in opposite directions; (5) both forces act on a different object. In a sense, the principle of Newton’s third law is not simply action equals reaction.

In this chapter, Feynman assumes that action equals reaction is a truth. (In Volume 2, Section 26.2, he discusses the failure of the law of action and reaction when two charged particles are moving on orthogonal trajectories.) A special case of action equals reaction is to consider a third particle that is not on the same line as the other two particles. Importantly, the “net” forces on the first two particles are neither equal nor opposite. Mathematically, the forces on the three particles, say A, B, and C, can be resolved into parts such that F_AB = -F_BA, F_BC = -F_CB, and F_AC = -F_CA. If there are four particles, then we expect ⁴C₂ = 6 combinations of particles that have corresponding components of mutual interaction that are equal in magnitude and opposite in direction.

Questions for discussion:

1. What are the simple cases of motion which can be analyzed by numerical methods and direct mathematical analysis?

2. What are the difficulties in solving the three-body problem?

3. How would you state the principle of Newton’s third law of motion?

The moral of the lesson: if the first particle exerts a force on the second particle and pushing it with a certain force, then the second particle will push on the first with an equal force and in the opposite direction; these two forces act in the same line.

References:

1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

2. Newton, I. (1687/1995). The Principia (translated by A. Motte). New York: Prometheus.

3. Šuvakov, M., & Dmitrašinović, V. (2013). Three classes of Newtonian three-body planar periodic orbits. Physical review letters, 110(11), 114301.

Section 9–7 Planetary motions

(Motion of a planet / Motion of Neptune, Jupiter, and Uranus / Computation time)

In this section, the three interesting points discussed are the motion of a planet, motions of Neptune, Jupiter, and Uranus, as well as the computation time for a numerical analysis.

1. The motion of a planet:

“The above analysis is very nice for the motion of an oscillating spring, but can we analyze the motion of a planet around the sun? (Feynman et al., 1963, section 9–7 Planetary motions).”

To be more precise, the Sun, Earth, and all planets in our solar system rotate around the barycenter (center of mass of all astronomical objects in the solar system). To achieve an approximation to the elliptical motion of a planet around the Sun, Feynman assumes that the Sun is infinitely heavy. Furthermore, we ignore how the Sun rotates about the center of our galaxy. Importantly, the Sun would move because of Newton’s third law of motion and gravitational forces on the Sun due to the planets. Essentially, we compute the motion of a planet orbiting around the Sun in an elliptical curve by using Newton’s second law of motion and Newton’s law of gravitation. In addition, we need to determine the “initial” position of a planet and how it is moving with a certain velocity.

To analyze the motion of a planet further, we have to calculate the components of the planet’s acceleration along two directions x and y. Feynman says that we shall suppose z is always zero because there is no force in the z-direction and there is no initial velocity v_z. Alternatively, one may assume that all planets in our solar system lie on the same plane. Physicists may explain that most of the mass of a solar system are on the same plane except for Pluto due to its unusual orbit. More importantly, Leapfrog method is used again in the simulation of a planet’s orbit as Feynman calculates the velocities v_x(0.05) and v_y(0.05) instead of the velocities v_x(0.00) and v_y(0.00).

2. The motion of Neptune, Jupiter, or Uranus:

“…let us see how we can calculate the motion of Neptune, Jupiter, Uranus, or any other planet (Feynman et al., 1963, section 9–7 Planetary motions).”

According to Feynman, we can calculate the force on planet number i, by using equations such as m_idv_ix/dt = − Gm_im_j(x_i−x_j)/r_ij³. Curiously, he specifies i = 1 to represent the Sun, i = 2, Mercury, i = 3 Venus, and so on. This is potentially misleading because the Sun is not a planet. Thus, he could have used the term “astronomical body i” instead of “planet number i.” Similarly, we can still use Leapfrog (midpoint) method to determine the motions of the n astronomical bodies including the Sun and planets. The positions of these astronomical bodies can be downloaded from NASA websites.

Unlike the previous two-body problem, this example has nine bodies (it seems that Pluto is excluded). Note that it is not even easy to solve a three-body problem. In general, a complete solution for the three-body problem would have the positions and velocities of the bodies for a period of time, provided three initial positions and initial velocities were determined. The motion of the three bodies is generally non-repeating and possibly “chaotic,” except in special cases. Interestingly, Feynman mentions in the next chapter that “the famous three-body problem, which so long challenged human powers of analysis.” The three-body problem is a special case of the n-body problem that is even more complicated.

3. Computation time:

“…we would need 4 × 10⁵ cycles to correspond to one revolution of a planet around the sun. That corresponds to a computation time of 130 seconds (Feynman et al., 1963, section 9–7 Planetary motions).”

Feynman explains that a very good computing machine may take a millionth of a second to do an addition. Furthermore, he estimates that it takes only two minutes to compute the motion of Jupiter around the Sun while including the perturbations of all the planets correct to one part in a billion. Currently, a computer can do multiple computations at the same time, and so the time needed for an addition is not a useful indicator of how fast the motions of planets can be computed. Specifically, the physical size of a processor, software program, and the speed of light are factors that determine the computation time needed. It should be worth mentioning that Feynman was involved in the “Connection Machine” project that connects 64000 processors for numerical computing (Hillis, 1989).

Feynman suggests that we can calculate motions of planets by using a computing machine to handle the arithmetic. He also claims that it is possible to achieve as high a degree of precision as we wish even for the tremendously complex motions of the planets. On the contrary, one may argue that there are dark matter particles (or unknown matters) in our solar system. Some physicists propose that dark matter have gravitational effects on motions of planets orbits despite difficulties in detecting the hypothetical matter (Gibney, 2017). However, the mass of astronomical bodies may not be sufficiently accurate and there could be a correction factor that accounts for “unknown matters” in our solar system in the first place.

Questions for discussion:

1. How would you compute the motion of a planet by using the Leapfrog method?

2. How would you compute motions of eight planets by downloading the initial positions of all planets?

3. What is the minimum computation time needed to simulate motions of planets in our solar system?

The moral of the lesson: we can compute complex motions of the planets to a relatively high degree of precision by using Leapfrog (midpoint) method.

References:

1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

2. Gibney, E. (2017). Dark-matter hunt fails to find the elusive particles. Nature News, 551(7679), 153.

3. Hillis, W. D. (1989). Richard Feynman and the Connection Machine. Physics Today, 42(2), 78-83.

Section 9–6 Numerical solution of the equations

(Euler’s method / Leapfrog method / Accuracy of numerical methods)

In this section, the three interesting points are Euler’s method, Leapfrog method, and the accuracy of numerical methods in calculating positions and velocities of an object.

1. Euler’s method:

“… we would find the motion only rather crudely because ϵ = 0.100 sec is rather crude, and we would have to go to a very small interval, say ϵ = 0.01 (Feynman et al., 1963, section 9–6 Numerical solution of the equations).”

Feynman chooses a time interval of 0.100 sec (step size, ϵ) and sets the initial position of an object as x(0) = 1.00. To calculate x(0.1), he adds the position x(0) by the velocity (which is zero) times 0.100 sec. The equations are x(t+ϵ) = x(t) + ϵv_x(t) and v_x(t+ϵ) = v_x(t) + ϵa_x(t) = v_x(t) − ϵx(t). This numerical method may be rewritten as y_n+1 = y_n + h ´ y’(y_n, t_n) and it is also known as Euler’s forward method (Zwillinger, 2014). In other words, this is an explicit method because y_n+1 is explicitly known in terms of y_n and y’(y_n, t_n). Although it is easy to implement the method, it has a problem of numerical stability depending on the step size chosen. Alternatively, Euler’s backward method is based on y_n+1 = y_n + h ´ y’(y_n+1, t_n+1) in which y’(y_n+1, t_n+1) is not known; this implicit method needs more computations, but it is relatively more stable.

Euler’s method uses the tangent line to the function at the beginning of the interval as an estimate of the slope of the function over the interval. This method can be derived by Taylor’s theorem, but it is sometimes criticized because it is not usually accurate for use in common applications (Burden & Faires, 2011). Essentially, the errors of this numerical method are dependent on the step size or time interval in the example provided. Thus, Feynman suggests using a very small interval, say ϵ = 0.010 sec, if the motion calculated is rather inaccurate. However, even when very small step sizes are used, errors are being accumulated over a large number of steps and the estimated value will likely diverge from the actual value.

2. Leapfrog method:

“Similarly, the velocity at this halfway point is the velocity at a time ϵ before (which is in the middle of the previous interval) plus ϵ times the acceleration at the time t (Feynman et al., 1963, section 9–6 Numerical solution of the equations).”

Feynman explains that we should use the acceleration halfway between two “times” (t−ϵ/2, t+ϵ/2) to deduce the velocity at a later time (t+ϵ/2). Similarly, the position later x(t+ϵ) is equal to the position before x(t) plus ϵ times the velocity at the time in the middle of the interval (t+ϵ/2). Mathematically, we can express the equations as x(t+ϵ) = x(t) + ϵv(t+ϵ/2) and v(t+ϵ/2) = v(t−ϵ/2) + ϵa(t). Simply phrased, this is a mid-point method that uses the speed between the “now” speed and the “then” speed at the end of an interval. The midpoint method is better because it approximates an integral by using the integrand at the midpoint (instead of “initial” position or “final” position). By using the midpoint between two successive points, there is a slight improvement in the accuracy of the numerical analysis.

The midpoint method is sometimes known as Leapfrog method. Leapfrog method is a second order method that is more accurate than Euler’s method, which is only first order. The order of accuracy of a numerical method indicates how errors decrease in the limit as the step size tends to zero. A method with an order of accuracy n and is known as a n-th order accurate method if there exists a C > 0 such that the error is less than Chⁿ in which h is the step size. Another important feature of Leapfrog method is that it preserves the amplitude of the simple harmonic motion within computation errors (round-off and truncation). This method is useful because it is relatively simple and has better numerical stability than Euler’s method.

3. Accuracy of numerical methods:

“The agreement is within the three significant figure accuracy of our calculation (Feynman et al., 1963, section 9–6 Numerical solution of the equations).”

By using a numerical method, it shows that an object starts from rest, first picks up a little upward (negative) velocity and it loses some of its distance. Notably, the agreement of this simulation with the exact mathematical solution of the equation of motion, x = cos t, is within the three significant figure accuracy. This is an impressive illustration of the power of Leapfrog numerical method because it involves simple calculations and gives relatively accurate results. We can simply use a spreadsheet (e.g., Google Sheets or Microsoft Excel) to simulate the motion of the object. A graph of simple harmonic motion of an object can be plotted almost instantly.

Currently, physics teachers may teach Runge–Kutta methods, which are a family of implicit and explicit iterative methods. In 1895, Carl Runge published a numerical method that is more elaborate than Euler’s method and is capable of greater accuracy. Runge’s idea was to develop approximate solutions based on improved formulas by using the midpoint and trapezoidal rules. In 1901, Martin Wilhelm Kutta published a paper that contains the famous Runge-Kutta methodology for solving ordinary differential equations. In general, Runge-Kutta method of order four requires four evaluations per step, whereas Euler’s method requires one evaluation.

Questions for discussion:

1. What are the advantages and disadvantages of Euler’s method?

2. What are the advantages and disadvantages of Leapfrog method?

3. How would you compare the accuracy of Euler’s method and Leapfrog method in simulating simple harmonic motion of an object?

The moral of the lesson: we can use Leapfrog (midpoint) method to illustrate the power of numerical analysis that simple calculations can give accurate results.

References:

1. Burden, R. L. & Faires, J. D. (2011). Numerical Analysis, 9th ed. Boston: Brooks/Cole.

2. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

3. Zwillinger, D. (2014). Handbook of Differential Equations. Orlando: Academic Press.