On the Relativity of a Non-accelerating Spring’s Elasticity Constant

G.R.Dixon, Sept. 23, 2013

Hooke was first to state that the force exerted by a resting, stretched/compressed spring is proportional to the amount of its deformation:

**F** = -k **Dx**.
(1)

k is the proportionality constant and is usually called the *elasticity* constant. In the Lorentz transformation the force on an object is *not* a matter of agreement among various inertial observers. Force in relativity is subject to a complicated transformation. If, viewed from inertial frame K, an object is momentarily at rest, then from the perspective of frame K’ the transformation simplifies to

F_{x}’ = F_{x},
(2)

F_{y}’ = F_{y} / g,
(3)

g = 1 / (1 – u^{2} / c^{2})^{1/2} and u being the speed of K’ relative to K.

Evidently the transformations in Eqs. 2 and 3 are common to *all* physical forces, including Lorentz (electromagnetic) and elastic forces.

Let us consider the consequences of the force transformation on the elasticity constant, k, in Hooke’s law.

Fig. 1 depicts a positively charged particle, q, at rest in frame K and attached to the end of a rubber band. We shall assume that the band’s mass is zero when it is unstretched. The particle experiences a repulsive force in the electrostatic field of an equal charge anchored at the origin of K.

Figure 1

The repulsive electrostatic force is equal and opposite to the force exerted by the rubber band. According to Hooke, the force exerted by the band is

F_{y} = -k Dy,
(4)

where Dy is the amount the band is stretched in the positive y-direction, and k is the band’s elasticity "constant".

Viewed from frame K’, traveling toward +x with speed u, the Lorentz transformation specifies that

Dy’ = Dy. (5)

Let us assume that the force exerted by the band is, from the perspective of K’,

F_{y}’ = -k’ Dy’ = -k’ Dy, (6)

where k’ is the spring "constant" in K’. From Eq. 3:

-k’ Dy = -k Dy / g. (7)

Evidently in this orientation,

k’ = k / g. (8)

Fig. 2 depicts the same band/charge system, oriented along the x-axis.

Figure 2

Once again, Hooke stipulates that

F_{x} = -k Dx. (9)

But now, viewed from K’, the band’s length and amount of stretch are different than in frame K:

Dx’ = Dx / g. (10)

Again assume that the force exerted by the band, in K’, is

F_{x}’ = -k’ Dx’. (11)

From Eqs. 2 and 10, then,

-k’ Dx’ = -k Dx = -g k Dx’, (12)

and thus

k’ = g k. (13)

In Eq. 8 the rubber band is "limper" when it moves perpendicular to its length. But in Eq. 13 the band is "stiffer" when it moves parallel to its length. Relativistically, the band’s elasticity "constant" is actually a function of (a) its orientation, and (b) its speed relative to an inertial observer.

Let us derive a few equations relating the views from frame K’ to those from K. For the orientation depicted in Fig. 1, we have for T, the kinetic energy of the system in K,

T = 0. (14)

Thus the complete energy in K is

E = U = k Dy^{2} / 2 (15)

where U is the band’s potential energy. If E = mc^{2}, then the system mass in K is

m = k Dy^{2} / 2c^{2} (16)

Now according to the Lorentz transformations, Dy’ = Dy, m’ = gm, E’ = gE. And as we previously derived, gk’ = k. Hence the potential energy in K’ is

U’ = k’ Dy’^{2} / 2 = k Dy^{2} / 2g = U / g. (17)

and

gU’ = U. (18)

For the kinetic energy in K’ we have

T’ = E’ – U’ = gE – U / g = gU – U / g (19)

and thus

T’ = U’ (g^{2} – 1). (20)

Furthermore,

m’c^{2} = T’ + U’ = g^{2} U’, (21)

and thus

m’ = g^{2} U’ / c^{2}. (22)

For the orientation in Fig. 2 we again have

T=0 (23)

and the complete energy in K is

E = U = k Dx^{2} / 2. (24)

The system mass in K is thus

m = k Dx^{2} / 2c^{2}. (25)

In this orientation Dx’ = Dx / g and again m’ = gm, E’ = gE. But now, as previously derived, k’ = gk. Hence the potential energy in K’ is

U’ = k’ Dx’^{2} / 2 = gk Dx’^{2} / 2 = k Dx^{2} / 2g = U / g. (26)

Again, therefore,

gU’ = U. (27)

For the kinetic energy in K’ we have

T’ = E’ – U’ = gE – U/g = gU – U/g (28)

and thus once again

T’ = U’(g^{2} – 1). (29)

Also, here again

m’c^{2} = T’ + U’ = g^{2} U’. (30)

Despite the many similarities between the "perpendicular" orientation (Fig. 1) and the "parallel" orientation (Fig. 2), it is worth emphasizing that the elasticity "constant" transformations are sensitive to a spring’s orientation. In the perpendicular case, we have gk’ = k. In the parallel case we have k’ = gk.