Section 20–1 Torques in three dimensions

(Torque vector / Transformation of torque / Axial vector)

In this section, Feynman discusses the concept of a torque vector, the transformation of the torque by rotation, and an axial vector.

1. Torque vector:

“…from Newton’s laws we see that we did not have to assume that the motion was in a plane; when we differentiate xp_y − yp_x, we get xF_y − yF_x, so this theorem is still right (Feynman et al., 1963, section 20–1 Torques in three dimensions).”

According to Feynman, if a rigid body is rotating in three dimensions, what we deduced for two dimensions is still applicable. For example, it is still true that xF_y − yF_x is the torque in the xy-plane, or the torque is “around the z-axis.” This torque is also equal to the rate of change of angular momentum, xp_y − yp_x, by using Newton’s second law of motion. Feynman’s explanation is not sufficient because one may prefer further discussion of the difference in sign in xF_y − yF_x. As a suggestion, it is good to let students derive the torque vector by resolving a force into two components, F_x and F_y. They should be able to deduce independently that the moments due to the two components are clockwise and anti-clockwise respectively, or vice versa.

Feynman cautions that one may get the wrong sign for a quantity if the coordinates are not handled in the right way. He clarifies that we may write τ_yz = zF_y – yF_z because a coordinate system may be either “right-handed” or “left-handed.” The handedness of a coordinate system or rotation is merely a convention in which we can assign clockwise to be the positive direction. To be more precise, if a screw is right-handed, it means that the screw is moved forward when it is rotated clockwise. Clockwise and anti-clockwise are also dependent on the perspective of an observer. An operational definition of handedness can be provided using Wu’s experiment on parity violation that is not based on a convention (See Chapter 52).

2. Transformation of torque:

“We wanted to get a rule for finding torques in new axes in terms of torques in old axes, and now we have the rule (Feynman et al., 1963, section 20–1 Torques in three dimensions).”

Feynman shows the transformation of a torque based on the coordinates of two systems that are related by x′ = xcos θ + ysin θ, y′ = ycos θ – xsin θ, z′ = z. Specifically, we can transform the torque in a new co-ordinate system rotated anti-clockwise by a angle θ using the following equations: τ_x′ = τ_xcos θ + τ_ysin θ, τ_y′ = τ_ycos θ − τ_xsin θ, τ_z′ = τ_z. Some may find Feynman’s method of direct substitution to find the torque in new axes unnecessary or lack of insights. A shorter and simpler method is to make use of the fact that the torque is a vector that is invariant under rotation. Although the torque can be expressed as a sum of vectors (e.g., τ_x′y′ = xF_y − yF_x), it is not a real vector that has all vector properties.

Feynman elaborates that a torque is a twist on a plane and it does behave like a vector. The torque vector is perpendicular to the plane of the twist and its length is proportional to the strength of the twist. To a certain extent, the three components of the torque transform like a real vector under rotation. As an alternative, we can simply transform a torque vector using a two dimensional rotation matrix: (i) first column: the point (1, 0) is rotated by an angle of θ anti-clockwise and moved to (cos θ, sin θ). (ii) second column: the point (0, 1) is rotated by an angle of θ anti-clockwise and moved to (−sin θ, cos θ). The rotation is a linear transformation and it can be expressed as R(x, y) = R[x(1, 0)+y(0, 1)] = xR(1, 0)+yR(0, 1).

3. Axial vector:

“Vectors which involve just one cross product in their definition are called axial vectors or pseudo vectors (Feynman et al., 1963, section 20–1 Torques in three dimensions).”

Feynman explains that axial vectors involve just a cross product (or vector product) in the definition and provides examples such as torque, angular momentum, angular velocity, and magnetic field. On the contrary, we have polar vectors such as coordinate, force, momentum, velocity, and electric field. In Chapter 52, Feynman adds that “a ‘vector’ which, on reflection, does not change about as the polar vector does, but is reversed relative to the polar vectors and to the geometry of the space; such a vector is called an axial vector (Feynman et al, 1963, section 52–5 Polar and axial vectors).” In a sense, the term vector product is a misnomer because it actually produces an axial vector. Feynman says that the torque is a vector, but it is not really a vector that it is loosely stated in books or websites.

Feynman elaborates that the cross product is very important for representing the features of rotation. He asks why the torque is a vector and then says that it is a miracle of good luck that we can associate a single axis with a plane, and thus we can associate a vector with the torque. Historically, the concept of torque as an axial vector is based on the Gibbs vector system (Chappell et al., 2016). However, Jackson (1999) identifies a dangerous aspect of vector notation and writes that “[t]he writing of a vector as a does not tell us whether it is a polar or an axial vector (p. 270).” Interestingly, the Clifford vector system includes the concept of wedge product and distinguishes the electric and magnetic field as a vector and bivector respectively (without the need of the axial vector).

Questions for discussion:

1. How would you define a torque vector?

2. How would you deduce the torque vector in another coordinate system that is rotated by an angle?

3. How would you define an axial vector?

The moral of the lesson: a torque vector is perpendicular to the plane of the twist, its length is proportional to the strength of the twist, and it behaves like an axial vector.

References:

1. Chappell, J. M., Iqbal, A., Hartnett, J. G., & Abbott, D. (2016). The vector algebra war: a historical perspective. IEEE Access, 4, 1997-2004.

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. Jackson, J. D. (1999). Classical Electrodynamics (3rd ed.). John Wiley & Sons, New York.

Section 19–4 Rotational kinetic energy

(Centrifugal force / Coriolis force / Centripetal force)

In this section, Feynman discusses the concept of centrifugal force, Coriolis force, and centripetal force from a perspective of work done on an object during rotation. His discussion of rotational kinetic energy is uncommon because he did not include typical examples such as an object rolling down a slope that has both translational kinetic energy and rotational kinetic energy. To be more precise, this section could be titled “Coriolis force” because this is his main focus. Instead of ending the chapter abruptly, there could be a mention of real-life examples relating to the Coriolis force such as long-range artillery trajectories, Foucault pendulum, or typhoons.

1. Centrifugal force:

“When we are rotating, there is a centrifugal force on the weights (Feynman et al., 1963, section 19–4 Rotational kinetic energy).”

According to Feynman, there is a centrifugal force on the weights when they are being rotated. An observer in an inertial frame of reference may elaborate that there is actually a tension (centripetal force) that pulls the weights. Feynman adds that the work we do against the centrifugal force ought to agree with the difference in the rotational kinetic energy of the weights. As a suggestion, one may explain that the rotational work should be done by a real force such as frictional force. In short, the centrifugal force exists as a fictitious (pseudo) force and it is a useful concept for an observer in a rotating frame of reference.

Feynman clarifies that the rotational work cannot be contributed by the centrifugal force because it is a radial force. This means that the centrifugal force is not the entire story in a rotating system. In chapter 12, Feynman says that “[a]nother example of pseudo force is what is often called ‘centrifugal force’ (Feynman et al., 1963, section 12–5 Pseudo forces).” However, physics teachers may derive the centrifugal force using (dr/dt)_fixed = (dr/dt)_rotating + w ´ r to help understanding the nature of forces in a rotating frame of reference in which r is a displacement vector (Thornton & Marion, 2004). In the derivation, we can identify three forces: (1) centrifugal force (proportional to w²), (2) Coriolis force (proportional to w), and (3) Euler force (proportional to dw/dt).

2. Coriolis force:

“Now let us develop a formula to show how this Coriolis force really works (Feynman et al., 1963, section 19–4 Rotational kinetic energy).”

Feynman explains that the Coriolis force has a very strange property: an object moving in a rotating system would appear to be pushed sidewise. For example, a person moving radially inward on a carousel would feel a sidewise force that is parallel to −2mω × v. We should clarify that the Coriolis force does arise from the motion of the object because the force is velocity-dependent and vanishes if it stops. In a sense, we may feel the existence of Coriolis force (or effect) when we walk radially inward or outward on a carousel. Importantly, the Coriolis force is a fictitious force that does not obey Newton’s third law (action = reaction) because it does not arise from an interaction between two objects.

Feynman suggests his students walk along the radius of a carousel whereby one has to lean over and push sidewise. (If we cannot counter this sidewise force with a frictional force of magnitude 2mωv, we would experience a tangential acceleration of 2ωv.) Feynman simply derives a formula of Coriolis force using τ = F_cr = dL/dt = d(mωr²)/dt = 2mωr(dr/dt), in which F_c is the Coriolis force. If we do not assume the angular speed to be constant, we can obtain mr²(dω/dt) that is known as the Euler force. However, this derivation does not help to understand the factor of 2 in 2mωv. One may prefer a physical explanation of the two equal components of Coriolis force using a general derivation of Coriolis force.

3. Centripetal force:

“This is simply the centripetal force that Moe would expect, having nothing to do with rotation (Feynman et al., 1963, section 19–4 Rotational kinetic energy).”

Unlike the previous example, an object is moving tangentially at a constant speed around the circumference of a carousel. In this example, Moe (rotating frame) observes the object moving at a velocity v_M at an instant, whereas Joe (inertial frame) sees it moving at a velocity v_J = v_M + ωr. One may add that the centripetal force is not a new or different force, but it may be a frictional force between the object and the carousel. We can provide an explanation of the direction of the Coriolis force using the cross product, but this is covered in the next chapter. Intuitively, the sidewise force is radially outward because of an inertial effect in which the object tends to be thrown out of the carousel.

Feynman explains the centripetal force on an object using the equation F_r = −mv_J²/r = −mω²r −mv_M²/r −2mv_Mω. Alternatively, from the perspective of an inertial observer, a person moving tangentially at a constant speed is essentially maintained by a frictional force (Morin, 2003). We can split the frictional force acting on the person walking on the carousel into three parts: F_friction = m(v_M + ωr)²/r = mv_M²/r + 2mv_Mωr + mω²r. The first term (mv_M²/r) is the additional frictional force on the person’s feet if he walks in a circle of radius r. The second term (2mv_Mωr) is the frictional force needed to counter the Coriolis force. The third term (mω²r) is the minimal frictional force needed that is equal to the centrifugal force due to the rotation of the carousel.

Questions for discussion:

1. How would you explain the nature of the Coriolis force?

2. How would you explain the Coriolis force for an object moving radially at a constant speed in a rotating system?

3. How would you explain the Coriolis force for an object moving tangentially (along a circumference) at a constant speed in a rotating system?

The moral of the lesson: the Coriolis force on an object is tangential when its velocity is radial, and the Coriolis force on the object is radial when its velocity is tangential.

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. Morin, D. (2003). Introductory Classical Mechanics. Cambridge: Cambridge University Press.

3. Thornton, S. T. & Marion, J. B. (2004). Classical Dynamics of Particles and Systems (5th Edition). Belmont, CA: Thomson Learning-Brooks/Cole.