Section 13–4 Gravitational field of large objects

(Infinite plane sheet / Thin spherical shell / Inside a sphere)

In this section, Feynman discusses how a gravitational field or potential energy varies due to an infinite plane sheet, thin spherical shell, and inside a sphere.

1. Infinite plane sheet:

“The gravitational field C at a mass point produced by an infinite plane sheet of matter (Feynman et al., 1963, section 13–4 Gravitational field of large objects).”

Feynman illustrates the validity of Gauss’ law for gravitation by calculating the gravitational field produced by an infinite plane sheet of material. He explains that the gravitational force is independent of distance because if we are closer, most of the matter is pulling at an unfavorable angle; if we are farther away, even more matter is situated more favorably to exert a pull toward the plane. Specifically, the force is weaker by the inverse square if we are farther away, but within the same cone of solid angle, it would be compensated by more matter that is proportional to the square of the distance. In other words, the farther away an object is from the infinite plane, more mass will be included within a “circle of influence” (or field of view with a cone of solid angle). Physics teachers could emphasize that Gauss’ law is applicable here because the gravitational force follows the inverse square law.

Gauss’ law for gravitation can be stated as: the flux of gravitational field through any surface is proportional to the total gravitational charge (mass). Similarly, we can use Gauss’ law to solve an electrical problem involving a large electrically charged plate that has an amount σ of charge per unit area. The electric field at any point outside the large plate is approximately equal to σ/2ϵ₀ in the outward direction and the approximation gets better if the plate is larger. If the plate is infinitely large, any point at a distance away from the plate is equivalent to every other point away from the plate. In Feynman Lectures on Computation, he adds that “defining a charge density here is important — it would be meaningless to discuss the total charge for an infinite plate (Feynman, 1996, p. 232).” This is because the total charge would be infinity in the idealized infinite plate.

2. Thin spherical shell:

“Thus, for a thin spherical shell, the potential energy of a mass m′, external to the shell, is the same as though the mass of the shell were concentrated at its center (Feynman et al., 1963, section 13–4 Gravitational field of large objects).”

We have assumed the gravitational force produced by the Earth at a point on the surface or outside it, is the same as if all the mass of the Earth were located at its center. This assumption is related to Newton’s shell theorem in which we conceptualize a spherical body as thin uniform hollow shells. Feynman’s approach is to work with the potential energy instead of the gravitational force because we do not have to worry about angles, that is, we merely add the potential energies of all the particles. However, one may expect Feynman to take the derivative of the potential energy (or at least mention the result) to obtain the gravitational force. This step is not really complicated and it may help students to appreciate Newton’s shell theorem better.

It may not be clear to students how the work done results in the potential energy due to a ring of mass is −Gm′dm/r. In a sense, Feynman partially fills up this gap only in the next chapter. In Chapter 14, he adds that “[t]he potential energy of gravitation for point masses M and m, a distance r apart is U(r) = −GMm/r (14.5). The constant has been chosen here so that the potential is zero at infinity (Feynman et al., 1963, section 14–3 Conservative forces).” Importantly, the work done by gravity results in negative gravitational potential energy because the gravitational force (not external force) is attractive and brings objects closer. The decrease in potential energy also implies an increase in the kinetic energy of the objects.

Lastly, Feynman only mentions the Earth acts as if all the materials were at the center by imagining it as a series of spherical shells. It is not trivial to explain how the gravitational force between two spherical objects is equivalent to the case when the mass of both objects are located at their center of mass. Young and Freedman (2004) explain that “the forces the two bodies exert on each other are an action-reaction pair, and they obey Newton’s third law. So we have also proved that the force that m exerts on the sphere M is the same as though M were a point. But now if we replace m with a spherically symmetric mass distribution centered at m’s location, the resulting gravitational force on any part of M is the same as before, and so is the total force. This completes our proof.” Notably, they have assumed that the distribution of mass remains spherical (or undistorted) in the presence of a massive object.

3. Inside the sphere:

“If the potential energy is the same no matter where an object is placed inside the sphere, there can be no force on it (Feynman et al., 1963, section 13–4 Gravitational field of large objects).”

Inside a spherical shell, the work done on an object by gravity, W = −Gm′m/a, is independent of R and independent of position. Therefore, no force is needed to move the object anywhere within the hollow sphere and no work is done when we move it about inside. It is similar to no work is done in moving the object horizontally on a flat table. (In short, W = −Gm′m/a Þ same potential energy within a shell Þ gravitational force, F = 0 within a shell.) If a point mass m is positioned at a distance r from the center of the Earth, we can determine the gravitational force by ignoring the spherical shells whose radius are longer than r; we only need to focus on the spherical shells whose radius are shorter than r.

Note: Physics teachers could compare the potential energy inside and outside a shell as shown below.

(i) Outside the shell: r_max – r_min = (R + a) – (R – a) = 2a

 Þ W = −Gm′m(r_max – r_min)/2aR = −Gm′m/R

(ii) Inside the shell: r_max – r_min = (R + a) – (a – R) = 2R

 Þ W = −Gm′m(r_max – r_min)/2aR = −Gm′m/a (constant)

The validity of Gauss’ law depends on the inverse square law of Coulomb. In volume II, Feynman provides a better explanation: “[i]f the force law were not exactly the inverse square, it would not be true that the field inside a uniformly charged sphere would be exactly zero. For instance, if the force varied more rapidly, like, say, the inverse cube of r, that portion of the surface which is nearer to an interior point would produce a field which is larger than that which is farther away, resulting in a radial inward field for a positive surface charge (Feynman et al., 1964, section 5.8 Is the field of a point charge exactly 1/r²?).” Thus, the net electric field is also zero for any two symmetric or opposite cones (same solid angle) diverging from the opposite sides of a point P as the fields due to them are in opposite direction (See Feynman et al., 1964, Fig. 5–9). We can apply the same logic for gravitational fields.

Questions for discussion:

1. How would you explain the gravitational field produced by an infinite plane sheet of material is constant or independent of distance?

2. Why is the gravitational force due to the Earth at a point on the surface or outside it, is the same as if all the mass of the earth were located at its center?

3. Why is the gravitational force equal to zero for any point within a spherical shell of mass?

The moral of the lesson: the gravitational force due to the Earth at a point on its surface or beyond, is the same as if all the mass of the Earth were located at its center, can be shown using Newton’s shell theorem.

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. Feynman, R. P., Leighton, R. B., & Sands, M. (1964). The Feynman Lectures on Physics, Vol II: Mainly electromagnetism and matter. Reading, MA: Addison-Wesley.

3. Feynman, R. P. (1996). Feynman lectures on computation. Reading, Massachusetts: Addison-Wesley.

4. Young, H. D., & Freedman, R. A. (2004). Sears and Zemansky’s University Physics (11th ed.). California: Addison Wesley.

Section 13–3 Summation of energy

(Time derivative of energy / Energy between the pairs / Sum of the works)

In this section, Feynman discusses the time derivative of energy, gravitational potential energy between any pairs of particles, and the sum of works done that brought in the particles.

1. Time derivative of energy:

“… we see that Eqs. (13.16) and (13.15) are precisely the same, but of opposite sign, so that the time derivative of the kinetic plus potential energy is indeed zero (Feynman et al., 1963, section 13–3 Summation of energy).”

Feynman moves on to a more general consideration of what happens to the total energy of a large number of particles. He suggests that we can prove the constancy of summation of energies by differentiating the kinetic energies of all particles and gravitational potential energies of all pairs of particles in a system. This is not a conventional way to show the total energy is a constant. Alternatively, one may explain that the total energy of a closed (isolated) system is constant. On the other hand, the total energy of an open system is not constant because it allows energy transfer. The constancy of summation of energies is dependent on how we define a system of particles.

In chapter 4, Feynman states that the law of conservation of energy is closely related to the symmetry of time in quantum mechanics. In his Ph.D. thesis titled A New Approach to Quantum Theory, he writes that “in the theory of action at a distance, the kinetic energy of the particles is not conserved. To find a conserved quantity, one must add a term corresponding to the ‘energy in the field’ (Feynman, 1942, p. 11).” He proceeds to show the constancy of total energy of particles (or Noether’s theorem) using a transformation of time-displacement. Based on an interview with Feynman, Mehra (1994) writes that “Feynman proved a thing called Noether’s theorem, not knowing that it was known (p. 132).”

2. Energy between the pairs:

“…the potential energy is also simple, it being also just a sum of contributions, the energies between all the pairs (Feynman et al., 1963, section 13–3 Summation of energy).”

Feynman explains that the involvement of potential energy is simple because it is only about the gravitational potential energy between all pairs of particles. In addition, the symbol S means that given values of i and j occur only once and we can keep track of this if we let i range over all values 1, 2, 3, …, and for each i, we can restrict the range of j to be greater than i. Simply phrased, one may specify the condition i < j (or state it as a subscript) to prevent a possibility of ‘double counting’ in listing the terms. It could be clearer if Feynman also writes down all the terms, for example, -Gm₁m₂/r₁₂ -Gm₂m₃/r₂₃ -Gm₁m₃/r₁₃, for a system of three particles.

Physics teachers could explain that the gravitational potential energy is a property of a system of particles instead of an individual particle. In other words, we cannot divide the total gravitational potential energy by specifying the quantity of energy that belongs to every particle. Some textbook authors elaborate that “if M >> m, as is true for Earth (mass M) and a baseball (mass m), we often speak of ‘the potential energy of the baseball.’ We can get away with this because, when a baseball moves in the vicinity of Earth, changes in the potential energy of the baseball – Earth system appear almost entirely as changes in the kinetic energy of the baseball, since changes in the kinetic energy of Earth are too small to be measured (Halliday, Resnick, & Walker, 2005).” In essence, it is inappropriate to specify the gravitational potential energy of individual object if the objects in a system are comparable in mass.

3. Sum of the works:

“Therefore the work done is the sum of the works done by each because if F₃ can be resolved into the sum of two forces, F₃ = F₁₃ + F₂₃… (Feynman et al., 1963, section 13–3 Summation of energy).”

We can understand the sum of works done (or change in potential energy of particles) as follows: Firstly, when we bring in the first particle, there is no work done because no other particles are present to exert a force on it. Secondly, when we bring in the second particle, there is only work done between the two particles, W₁₂ = −Gm₁m₂/r₁₂. Thirdly, when we bring in the third particle, the work done is the sum of the works done against the first two particles because F₃ can be resolved as a sum of two forces, F₃ = F₁₃ + F₂₃. Mathematically, the net work done can be written as ½(−Gm₁m₂/r₁₂ −Gm₂m₃/r₂₃ −Gm₁m₃/r₁₃) and it is independent of the order of the particles that are brought to their final positions. However, one may explain that the potential energy of particles is stored in the fields instead of sharing among the particles.

In the case of electrostatic energy, it can be expressed as “U = ½[q₁ϕ(1)+q₂ϕ(2)]… That is why we need the factor ½ (Feynman et al., 1964, Chapter 8).” Using Feynman’s notations, the gravitational potential energy stored in three objects can also be written as U = ½[m₁Y(1) + m₂Y(2) + m₃Y(3)]. Initially, Feynman explains that the idea of locating the energy in the field is inconsistent with the assumption of the existence of point charges. To resolve this inconsistency, he suggests that “[o]ne way out of the difficulty would be to say that elementary charges, such as an electron, are not points but are really small distributions of charge… (Feynman et al., 1964, Chapter 8).” One may use the principle of mass-energy equivalence to argue that the gravitational field has energy, and thus it has mass.

Questions for discussion:

1. What is/are the implication(s) if the time derivative of the kinetic plus potential energy is zero?

2. How would you explain the gravitational potential energy is a property of a system of particles?

3. How would you explain the work done is the sum of the works done by each force?

The moral of the lesson: we can write the potential energy as a sum over each pair of particles because gravitational forces obey the principle of superposition of forces.

References:

1. Feynman, R. P. (1942/2005). Feynman’s thesis: A New Approach to Quantum Theory. Singapore: World Scientific.

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. Feynman, R. P., Leighton, R. B., & Sands, M. (1964). The Feynman Lectures on Physics, Vol II: Mainly electromagnetism and matter. Reading, MA: Addison-Wesley.

4. Halliday, D., Resnick, R., & Walker, J. (2005). Fundamentals of Physics (7th ed.). New York: Wiley.

5. Mehra, J. (1994). The Beat of a Different Drum: The life and science of Richard Feynman. Oxford: Oxford University Press.

Section 13–2 Work done by gravity

(One-dimensional case / Any closed path / A simple triangle)

In this section, Feynman discusses work done by gravity from the perspectives of a one-dimensional case, any closed path, and a simple triangle.

1. One-dimensional case:

“This one-dimensional case is easy to treat because we know that the change in the kinetic energy is equal to the integral (Feynman et al., 1963, section 13–2 Work done by gravity).”

In a simple example, an object starts from a point and moves radially toward the Sun or the Earth. By using a convention in which moving radially outward is positive, the gravitational force is −GMm/r² and thus, the object’s gravitational potential energy decreases (or kinetic energy increases) as the distance fallen increases. One may ask if it is possible to find another formula for gravitational potential energy that is different from mgh such that the law of conservation of energy is still true. In Tips on Physics, Feynman (2006) replies that “the potential energy between two particles is −Gm₁m₂/r. It’s hard for me to remember which way the potential energy goes. Let’s see: the particles lose energy when they come together, so that means when r is smaller, the potential energy should be less, so it’s negative-I hope that’s right! I have a great deal of difficulty with signs (p. 48).” However, the sign of a work done and change in potential energy is also related to how we define a system.

The word system refers to one or more objects that we mentally group together as a matter of convenience, but it may affect how we classify forces into internal or external. For example, the gravitational force is an “internal” force instead of an external force (e.g., your hand) that does the work in the object-Earth system. The work done by an internal force results in a decrease in potential energy and it is commonly expressed as W = -DU. On the contrary, our hand (external force) can lift an object such that there is an increase in potential energy. In general, it is imprecise to simply write “potential energy of an object”; the potential energy belongs to a system which may consist of “the object and the Earth” (Lindsey et al., 2012). Similarly, weight is not a property of an object, but it is a measure of the gravitational force between the object and the Earth (Serway & Faughn, 2003).

2. Any closed path:

“…there is no friction the object should come back with neither higher nor lower velocity—it should be able to keep going around and around any closed path (Feynman et al., 1963, section 13–2 Work done by gravity).”

Feynman asks whether it is possible to make perpetual motion in a gravitational field by using a fixed, frictionless track. Some students may guess that it does not seem possible because the gravitational force varies in different locations and has different strengths and in different directions. Feynman suggests that it is impossible to have perpetual motion, but we ought to find out if this is indeed the case. Recently, Wilczek (2012) proposes the concept of time crystal and explains that “a system with spontaneous breaking of time translation symmetry in its ground state must have some sort of motion in its ground state, and is therefore perilously close to fitting the definition of a perpetual motion machine (p. 1).” In short, it is possible to have “almost perpetual motion” just like the electric current in a superconductor.

On the question “is the work really zero?” Feynman answers that it is zero and then examines it mathematically by using a simple path as shown in Fig. 13–3. That is, a small object is carried from point 1 to point 2, and then it is made to go around a circle to point 3, back to 4, then to 5, 6, 7, and 8, and finally back to point 1. The total work done by a gravitational force in moving around this path or any closed path is zero. Physics teachers may elaborate that this is true for a conservative force (it is discussed later in chapter 14). Generally speaking, examples of conservative forces include gravitational forces and electrical forces, whereas examples of non-conservative forces include frictional forces and air resistance. Some physicists may add that the work done is definitely zero provided it satisfies the condition: curl of F = 0.

3. A small triangle:

“…we see that the work done in going along the sides of a small triangle is the same as that done going on a slant because scos θ is equal to x (Feynman et al., 1963, section 13–2 Work done by gravity).”

In this example, Feynman discretizes a curve into a series of sawtooth jiggles as shown in Fig. 13–4. Importantly, each triangle has to be very small such that the (variable) gravitational force is approximately constant over the entire triangle. Thus, the work done in going along the sides (horizontal and vertical) of a small triangle is numerically the same as the work done in going on a slant side because s ´ cos θ is equal to x. Notably, the “smooth” closed path can be approximated by a series of radial and circumferential steps. In a sense, Fig. 13.4 is misleading because some students may think that there are only horizontal and vertical steps instead of radial and circumferential steps.

There are several issues in Feynman’s example of a mass on a vertical spring. If we pull the mass downward, there is a change in the elastic potential energy as well as the gravitational potential energy. Thus, physics teachers should simplify the example by discussing a horizontal spring that slides on a frictionless table instead of a vertical spring. Furthermore, the work done can be distinguished as “work done on a spring” and “work done by a spring.” For instance, if your hand pulls a mass (or extends a spring), the work done on a spring by your hand (external force) is (+kx)(+x) and positive. At the same time, the work done by the spring or internal force is (–kx)(+x) and it is negative. As a suggestion, physics teachers may end the section by explaining that we have usually ignored the work done in compressing an (non-rigid) object while pushing it.

Questions for discussion:

1. How would you explain the work done and a change in potential energy in a one-dimensional case?

2. Why is the total work done by a gravitational force in moving any closed loop is zero?

3. Why is the work done in moving along two sides of a small triangle is the same as the work done in moving on a slant side?

The moral of the lesson: the total work done by a gravitational (conservative) force in moving any closed path is zero.

References:

1. Feynman, R. P., Gottlieb, & M. A., Leighton, R. (2006). Feynman’s tips on physics: reflections, advice, insights, practice: a problem-solving supplement to the Feynman lectures on physics. San Francisco: Pearson Addison-Wesley.

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. Lindsey, B. A., Heron, P. R. L., & Shaffer, P. S. (2012). Student understanding of energy: Difficulties related to systems. American Journal of Physics, 80(2), 154-163.

4. Serway, R. A., & Faughn, J. S. (2003). College Physics (6th ed.). Pacific Grove, California: Brooks/Cole.

5. Wilczek, F. (2012). Quantum time crystals. Physical review letters, 109(16), 160401.

Section 13–1 Energy of a falling body

(Rate of change of potential energy / Power expended / Work done by gravity)

In this section, Feynman discusses the rate of change of potential energy, power expended, and work done by gravity.

1. Rate of change of potential energy:

“…the rate of change of the kinetic energy is −mg(dh/dt), which quantity, miracle of miracles, is minus the rate of change of something else! It is minus the time rate of change of mgh! (Feynman et al., 1963, section 13–1 Energy of a falling body).”

Feynman did not begin chapter 13 by discussing the work-kinetic energy theorem. Firstly, he derives how the time rate of change of kinetic energy is related to the time rate of change of potential energy. Next, he specifies that the gravitational force is constant and it is equal to −mg. This is an approximation because we can show that GMm/(R + h)² » mg if the height h is less than 100 km above the sea level. In a sense, this section (13.1) can be titled as “work done by constant gravitational force, whereas the next section (13.2) is about “work done by variable gravitational force.” Furthermore, we assume there is no air resistance (an idealization) such that there is a miracle of miracles (Feynman’s words) in which the rate of change of the kinetic energy is minus the rate of change of potential energy.

In Tips on Physics, Feynman (2006) elaborates that “the value of the potential energy can be zero any place you want. The way we’re going to use potential energy is to talk about its changes - and then, of course, it doesn’t make any difference if you add a constant (p. 47).” In other words, what matters is the change in gravitational potential energy rather than its absolute value. (There are locations, e.g., sea level or somewhere at infinity, where they can be set to zero as a matter of convenience.) In this section, Feynman proves that the work done by gravity results in a change of gravitational potential energy. This is in contrast to conventional physics textbook authors that prove the work done by a force results in a change of kinetic energy. Nevertheless, one should appreciate the beauty of how the time rate of change of kinetic energy, −mg(dh/dt), is the minus time rate of change of mgh (potential energy).

2. Power expended:

“We thus have a marvelous theorem: the rate of change of kinetic energy of an object is equal to the power expended by the forces acting on it (Feynman et al., 1963, section 13–1 Energy of a falling body).”

Using vector analysis, Feynman shows that dT/dt = F_xv_x + F_yv_y + F_zv_z. = F.v. He also expresses F.v as the power being delivered to an object: the dot product of a force acting on an object and the velocity of the object. Moreover, he describes a theorem: the rate of change of kinetic energy of an object is equal to the power expended by the forces acting on it. Weaker students may prefer Feynman adds one more step such that d(½mv²)/dt = d(½mv.v)/dt = ½m(v.dv/dt + dv/dt.v) = m(dv/dt).v = F.v. Conversely, some textbook authors simply state dT/dt = F.v. To be precise, physicists prefer using the term instantaneous power instead of power.

In daily lives, the word power may mean average power. It is worthwhile to distinguish average power and instantaneous power. Essentially, average power (ΔW/Δt) is the amount of work done (ΔW) by a force divided by a period of time (Δt). In general, if the force acting on an object is constant and the average velocity is v_ave, then the average power can be shown to be equal to F.v_ave. If the object’s initial velocity is zero, then it is possible that the average power is equal to F.v_max/2 in which v_max is the maximum velocity of the object. On the other hand, the instantaneous power is numerically equal to the average power as the time taken, Δt, approaches zero. Mathematically, the time rate of change of energy transferred is equal to the dot product of the force at an instant and the instantaneous velocity, F.v.

3. Work done by gravity:

“We also see that it is only a component of force in the direction of motion that contributes to the work done (Feynman et al., 1963, section 13–1 Energy of a falling body).”

The work done by a force on an object can be written as F.s or F_xdx + F_ydy + F_zdz. In a daily life example, the gravitational force can be vertical, and it has only a single component, say F_z, that is equal to –mg. (Based on the convention that upward is positive, the gravitational force is acting downward and thus, it is negative.) Using this example, Feynman shows that ∫²₁F.ds = −mg(z₂ − z₁). In short, this is equivalent to the work-potential energy theorem, DW = DP.E. (or DW = −DP.E. for gravity). For instance, if your hand lifts a book upward, the work done by your hand (external force) is (+mg)(+h) and it is positive. At the same time, the work done by the gravitational force is (–mg)(+h) and thus, negative. Because of Newton’s third law of motion, we have two different works that are equal in magnitude and opposite in sign.

In Tips on Physics, Feynman (2006) explains that “[i]n certain cases, that integral can be calculated easily ahead of time, because the force on the particle depends only on its position in a simple way. Under those circumstances, we can write that the work done on the particle is equal to the change in another quantity called its potential energy, or P.E. Such forces are said to be ‘conservative’: DW = DP.E. (p. 44).” Strictly speaking, we need to idealize the particle moves slower than a snail (its speed is infinitesimally close to zero) such that there is no gain in kinetic energy. Some may prefer a more realistic work-energy theorem that includes both potential energy and kinetic energy: DW = DP.E. + DK.E.

Questions for discussion:

1. How do you explain the time rate of change of kinetic energy, −mg(dh/dt), is the minus time rate of change of mgh (potential energy)?

2. How do you prove that the rate of change of kinetic energy of an object is equal to the power expended by the forces acting on it?

3. How do you show that the work done by a force is positive or negative?

The moral of the lesson: average power is equal to work done per second and it is only a component of force in the direction of motion that contributes to the work done.

References:

1. Feynman, R. P., Gottlieb, & M. A., Leighton, R. (2006). Feynman’s tips on physics: reflections, advice, insights, practice: a problem-solving supplement to the Feynman lectures on physics. San Francisco: Pearson Addison-Wesley.

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.

Section 12–6 Nuclear forces

(Interaction range / Mathematical approximation / Fundamental machinery)

In this section, Feynman discusses nuclear forces from the perspectives of interaction range, mathematical approximation, and fundamental machinery.

1. Interaction range:

“These forces have a very tiny range which is just about the same as the size of the nucleus, perhaps 10⁻¹³ centimeter (Feynman et al., 1963, section 12–6 Nuclear forces).”

Feynman concludes this chapter with a brief discussion of nuclear forces. He mentioned that there was no known law of nuclear forces. Currently, it is still challenging to accurately calculate the force between two nuclei. For instance, Wilczek (2007) writes that “[i]n principle, the equations of QCD contain all the physics of strong internucleon forces. But in practice, it is extremely difficult to solve the equations and calculate those forces (p. 156).” The physical law of nuclear forces is now known as quantum chromodynamics (QCD) and it is still true that these forces have a very tiny range which is about 10⁻¹³ centimeter (or about the size of a nucleon). One may criticize the use of the term nucleus here because it is possible to be much larger than a nucleon.

Physicists, nowadays, use the term “strong force” instead of “nuclear force.” We may estimate the range of strong force to be about the size of a nucleon because the quarks enclosed are point-like objects. Wilczek (2015) explains that “[t]he use of ‘strong’ and ‘force’ together is potentially ambiguous because ‘strong force’ could be taken to mean a powerful source of acceleration. Thus when speaking of the gravitational influence of a neutron star, or a black hole, one might say that gravity exerts a strong force on a nearby planet. To avoid ambiguity, in such cases I use terms like ‘powerful force’ or ‘powerful interaction,’ avoiding ‘strong force’ and ‘strong interaction’ (p. 389).” Strictly speaking, the concept of force is also not suitable in QCD (or quantum theory), but the concept of energy is more natural.

2. Mathematical approximation:

“Any formula that can be written for nuclear forces is a rather crude approximation which omits many complications… (Feynman et al., 1963, section 12–6 Nuclear forces).”

Feynman mentions that any formula that can be written for nuclear forces is only a crude approximation which neglects many complications. He adds that nuclear forces disappear as soon as the particles are at a great distance apart, and states that they are very strong within the 10⁻¹³ cm range. Physicists currently explain that the nuclear force is a residual effect of the strong force. In short, the strong force is relatively stronger as the quarks separate from each other, and it is weaker if the quarks approach one another. The idea that strong force becomes weaker at short distances is known as asymptotic freedom (or “charge without charge”).

According to Feynman, nuclear forces do not vary inversely as the square of the distance, but they decrease exponentially over a distance r according to the equation, F = (1/r²)exp(−r/r₀), in which the distance r₀ is of the order of 10⁻¹³ centimeter (or 10⁻¹⁵ m). Mathematically, this is incorrect because it suggests that the nuclear forces approach infinity when the distance r approaches zero. Based on experiments of the deep inelastic scattering of electrons off nucleons, physicists deduce that quarks behave like free particles in the asymptotic limit of zero separation. Interestingly, Zee (2015) explains that “[q]uarks are like some lovers: When they are far apart, they want each other, but when they are close together, they barely acknowledge (p. 205).”

3. Fundamental machinery:

“…we do not understand them in any simple way, and the whole problem of analyzing the fundamental machinery behind nuclear forces is unsolved (Feynman et al., 1963, section 12–6 Nuclear forces).”

Feynman says that the laws of nuclear force are very complex and he does not have a simple way of understanding these laws. (He mentioned initially that there was no known law of nuclear forces.) The problem of analyzing the fundamental machinery of nuclear forces was unsolved when this lecture was delivered. George Zweig, Feynman’s doctoral student, developed a model and named the fundamental particles as “aces” in 1964. Gell-Mann also proposed a quark model within the same year. In a public lecture on QED, Feynman (1985) expresses that “Murray Gell-Mann nearly went crazy trying to figure out the rules by which all these particles behave, and in the early 1970s they came up with the quantum theory of strong interactions (or quantum chromodynamics), whose main actors are particles called ‘quarks’ (p. 132).”

Just like the electromagnetic force is mediated by photons, the strong force is mediated by gluons. During the public lecture on QED, Feynman (1985) explains that “[s]omething else has been invented to go back and forth and hold quarks together; something called ‘gluons’ (p. 134).” The gluons are labeled different colors (red, blue, and green) that behave like an electric charge. The color charge is conserved in the interactions of quarks such that the color of a quark is changed after absorption of a color gluon. However, Feynman (1985) responds that “[t]he idiot physicists, unable to come up with any wonderful Greek words anymore, call this type of polarization by the unfortunate name of ‘color,’ which has nothing to do with color in the normal sense (p. 136).”

Questions for discussion:

1. How would you explain the strong force is a short range force?

2. How would you justify the word “strong” as used in strong force?

3. How would you describe the fundamental machinery behind the strong force?

The moral of the lesson: we may understand the nature of strong force from the perspectives of its interaction range, relative strength, and the interactions of quarks.

References:

1. Feynman, R. P. (1985). QED: The strange theory of light and matter. Princeton: Princeton University Press.

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. Wilczek, F. (2007). Particle physics: Hard-core revelations. Nature, 445(7124), 156-157.

4. Wilczek, F. (2015). A Beautiful Question: Finding Nature’s Deep Design. New York: Penguin Press.

5. Zee, A. (2015). Fearful Symmetry: The Search for Beauty in Modern Physics. Princeton, NJ: Princeton University Press.