Models of forces

where the transformed arguments of the ϕij have been pointed out symbolically.

An elementary simplification yields if and only if ϕ_ij depends only on the vector differences x_k− x_l. Second, setting r = 0, R = 1 and t₀ = 0, G reduces to the translation by V of all the vector velocities. Then ϕ_ij(G(x⁰, v⁰, s⁰)) = ϕ_ij(x⁰, v⁰, s⁰) if and only if ϕ_ij depends only on the vector differences v_k− v_l, Third, setting t₀ = 0, and taking into account the previous results, G reduces to the rotation by R of all the vector differences x_k− x_l and v_k− v_l. Then ϕ_ij(G(x⁰, v⁰, s⁰)) = ϕ_ij(x⁰, v⁰, s⁰) if and only if ϕ_ij depends only on the moduli of the such vector differences. Finally, setting r = 0, V = 0 and R = 1, G reduces to the translation by t₀ of s⁰. The invariance of the ϕ_ij implies that they do not depend on s⁰.

Corollary 1.1. For an isolated particle (n = 1) the relativity principle implies f ≡ 0.

Proof. It is left as an exercise. Hint: if u ∈ R^d is such that Ru = u for any rotation matrix R, then u ≡ 0 (why?).

1.3 Models of forces

1.3.1 Interaction models

All classical physics (and a large part of the quantum one) is built up by a further phenomeno-logical simplification: the force scalars ϕij are supposed to be independent of the velocities and

1.3. MODELS OF FORCES 13 to depend only on the distance |x_i− x_j| of particles i and j: ϕ_ij = ϕ_ij(|x_i− x_j|). This is due to the experimental fact that two out of the four fundamental interactions of nature, namely the gravitational and the electrostatic one, share such a property (within certain limits). Thus the internal forces are assumed, in general, of the so-called central form

f_ij = ϕ_ij(|x_i− x_j|) x_i− x_j

|x_i − xj| = ϕ_ij(|x_i− x_j|)ˆx_ij , (1.12) where the unit vector ˆx_ij := _|x^xi−xj

i−xj| from j to i has been defined. The 3rd principle requires of course ϕij(r) = ϕji(r).

Exercise 1.2. Let Φ_ij(r) be minus a primitive function of ϕ_ij(r), i.e. Φ_ij(r) = −R ϕij(r)dr, or Φ⁰_ij(r) = −ϕ_ij(r). Show that

f_ij = ϕ(|x_i− x_j|) x_i− x_j

|x_i− xj| = − ∂

∂x_iΦ_ij(|x_i− x_j|) , (1.13) where ∂/∂x_j = ∇_xj denotes the gradient with respect to the components of the vector x_j. Show that ϕ_ij = ϕ_ji implies that f_ji = −∂Φ_ij/∂x_j.

Example 1.1 (Elastic force). The simplest possible assumption is that the force scalar be a linear homogeneous function, i.e. ϕ_ij(r) = −k_ijr, or ϕ_ij(|x_i − x_j|) = −k_ij|x_i − x_j|, where k_ij > 0 is a given constant. Then

f_ij = −k_ij(x_i− x_j) , (1.14)

i.e. the so-called elastic force, or Hooke’s law. Notice that k_ij > 0 implies that such a force is always attractive. This force models the small motions of a given system around its equilibrium positions. Remark: the 3rd principle implies k_ij = k_ji.

Example 1.2 (Gravitational force). For the gravitational force, Newton deduced the form ϕij(r) = −Gmimj/r², where mi and mj are the masses of particles i and j, respectively, whereas G = 6.67 10⁻⁸ cm³/(gs²) is the so-called gravitational constant, a fundamental con-stant of nature, like the speed of light, the Planck concon-stant, and so on. Then

f_ij = −G m_im_j

|x_i− x_j|³(x_i− x_j) . (1.15) The Newton force satisfies the 3rd principle and is always attractive.

Example 1.3 (Electrostatic force). For the electrostatic force between charged particles, Coulomb deduced the form ϕ_ij(r) = kq_iq_j/r², where q_i and q_i are the electric charges of particles i and j, respectively, whereas the positive constant k depends on the system of units adopted: k = 1 in the CGS system (with charges measured in esu, the elementary charge being 4.8 10⁻¹⁰esu).

Thus

f_ij = k q_iq_j

|x_i− x_j|³(x_i− x_j) . (1.16) The Coulomb force satisfies the 3rd principle and is attractive for charges of opposite sign, and repulsive for charges of equal sign.

The form of the Newton force (1.15) can be “deduced” in different ways. A simple and direct one is the following. Newton was aware of the experiments of Boyle on falling bodies in vacuum tubes, from which it follows that the weight force on a particle is proportional to its mass:

m¨x = −mgˆe₃, where ˆe₃is the upward unit vector, and g = 980 cm/s²is the gravity acceleration, implies ¨x = −gˆe₃, i.e. a falling law of the form x(t) = x(0) + v(0)t − ¹₂gˆe₃t². The fundamental intuition of Newton is that the force law ruling falling bodies and, neglecting the resistance of air, the motion of a cannon ball, is an approximate form of the force law ruling the planetary motions. For the latter, assuming a central force law of the form m¨x = GM_Omϕ(|x|)ˆx, where M_O is the mass of a star (e.g. the Sun) placed approximately at rest in the origin. The proportionality to m is the one of the falling body problem, that to M_O is due to the 3rd principle; at this level G is a dimensional proportionality constant. The other experimental information available to Newton was the third Kepler law: the ratio of the square of the revolution period T to the cube of the major semi-axis a of the (elliptic) planetary orbits is a constant, say 1/k, independent of the planet: T²/a³ = 1/k. Thus by rescaling the space and time variables in the Newton equation, measuring length in units of a and time in units of T , i.e. setting x = ax⁰, t = T t⁰, one has to get one and the same orbit, with unit period and unit semi-axis. The rescaling transforms the Newton law of the problem into

a T²

d²

dt⁰²x⁰ = GMOϕ(a|x⁰|)ˆx⁰ .

Multiplying the latter by a², and taking into account that a³/T² = 1/k, one gets d²

dt⁰²x⁰ = (kGM_O) a²ϕ(a|x⁰|)ˆx⁰ . (1.17) Now, imposing that the right hand side of the latter equation does not depend on the particular semi-axis a, i.e. requiring that the equation itself transforms to d²x⁰/dt⁰² = (kGM_O)ϕ(|x⁰|)ˆx⁰, yieldsa²ϕ(a|x⁰|) = ϕ(|x⁰|), or

ϕ(a|x⁰|) = a⁻²ϕ(|x⁰|) . (1.18)

The latter condition tells that ϕ(r) is a homogeneous function of degree −2, accoridng to the following definition.

Definition 1.1. A function f : R^d → R is said to be homogeneous of degree s ∈ R if f(λx) = λ^sf (x) for any λ > 0.

Theorem 1.1 (Euler’s theorem on homogeneous functions). If f : R^d → R is a homogeneous function of degree s, then x · ∇f (x) = sf (x). In particular, if d = 1, f (x) = c|x|^s, where c is a constant.

Proof. Taking the derivative of f (λx) = λ^sf (x) with respect to λ, and setting λ = 1, one gets x · ∇f (x) = sf (x). If d = 1, the latter equation is an ordinary differential equation for f , namely f⁰ = sf /x. By separating variables, one can rewrite it as df /f = sdx/x, which integrated gives ln |f | = s ln |x| + c⁰, i.e. ln(|f |/|x|^s) = c⁰. Taking the exponential and setting c = ±e^c0 on gets the result.

1.3. MODELS OF FORCES 15 By Euler’s theorem, relation (1.17) gives ϕ(r) = cr⁻²(Notice that the argument of ϕ is positive).

Thus the Kepler 3rd law implies that the force is inversely proportional to the square of the distance of the particle from the center of attraction (the star, or planet). The sign of the constant is then fixed by requiring that the force is attractive, so that bodies close to the surface of the Earth, move downwards. The unification of the two phenomena (planetary motion and falling bodies) gives g = GM_T/R²_T, where M_T and R_T are the mass and the radius of the Earth, respectively.

Remark 1.2. The method just sketched is the first example of scale invariance of a physical law. The Newton gravitation law, with the known force law, is invariant under the re-scaling x = λx⁰, t = λ^3/2t⁰ for any (dimensionless) λ > 0. Notice that |x|³/t² = |x⁰|³/t⁰², which is another (more general) way of stating the 3rd Kepler law. In some texts scale-invariance is referred to as “mechanical similarity” [2, 25].

Exercise 1.3. Consider the d-dimensional (d = 1, 2, 3) harmonic motion, defined by the New-ton law m¨x = −kx, x ∈ R^d. Rescale space and time variables like x = λx⁰ and t = λ^αt⁰, and determine the value of the scaling exponent α such that the Newton law is invariant under the given scaling transformation. Discuss the physical meaning of the result.

1.3.2 Drag forces

Taking into account the resistance exerted by a medium (typically a fluid or a gas) on body motions is a difficult problem. For example, a body moving in an ideal (i.e. collisionless) gas is subject to a force opposing its motion which is due to the collisions with the particles of the gas hitting on it. For a sphere of radius r and mass much larger than that of the gas particles, which moves with an instantaneous velocity v much higher than the thermal velocity of the gas particles (which is proportional to the square root of the gas temperature), the average force opposing to the motion, or drag force, is found to be [9]

f = −2πρr²

3 |v|v , (1.19)

where ρ is the mass density of the gas (mass per unit volume). The law (1.19) is also due to Newton.

On the other hand, the drag force exerted by a fluid on a sphere moving in it with a small instantaneous velocity v, is given by the Stokes law

f = −(6πρνr)v (1.20)

where ν is the kinematic viscosity coefficient of the fluid [1, 27]. The Stokes law is valid if |v|  ν/r. Both laws (1.19) and (1.20) admit corrections, to lower and higher velocities, respectively, whose derivations are not trivial. The final result can be resumed in a phenomenological formula for the drag force of the form

f = −k(|v|)v , (1.21)

where the drag coefficient k admits an expansion of the form k(|v|) = γ + k₁|v| + k₂|v|² + · · · at small velocities, with coefficients depending on the form of the moving body, on the density of the medium and so on. Formula (1.21) holds for both fluids and gasses (though for fluids a less simple dependence of k on |v| holds). Taking into account the dependence of k on v in applications is difficult. By far the most widespread form of drag force in physics assumes a constant drag coefficient, i.e. is of the form f = −γv.

Exercise 1.4. Consider a particle of mass m falling in a gas subject to its weight force. Assume a drag coefficient of the form k = γ + k₁|v|. Assume the vertical axis oriented downwards, so that in the falling motion v(t) > 0. The Newton equation of motion of the particle is m ˙v = −(γ + k₁v)v + mg (why?). Solve the equation with the initial condition v(0) = 0.

Question: which is the limit velocity v∗ := lim_t→∞v(t)? Hint: determine v∗ before solving the Newton equation.