Derivation of Classical Mechanics in an Energetic Framework via Conservation and Relativity

Abstract

The notions of conservation and relativity lie at the heart of classical mechanics, and were critical to its early development. However, in Newton’s theory of mechanics, these symmetry principles were eclipsed by domain-specific laws. In view of the importance of symmetry principles in elucidating the structure of physical theories, it is natural to ask to what extent conservation and relativity determine the structure of mechanics. In this paper, we address this question by deriving classical mechanics—both nonrelativistic and relativistic—using relativity and conservation as the primary guiding principles. The derivation proceeds in three distinct steps. First, conservation and relativity are used to derive the asymptotically conserved quantities of motion. Second, in order that energy and momentum be continuously conserved, the mechanical system is embedded in a larger energetic framework containing a massless component that is capable of bearing energy (as well as momentum in the relativistic case). Imposition of conservation and relativity then results, in the nonrelativistic case, in the conservation of mass and in the frame-invariance of massless energy; and, in the relativistic case, in the rules for transforming massless energy and momentum between frames. Third, a force framework for handling continuously interacting particles is established, wherein Newton’s second law is derived on the basis of relativity and a staccato model of motion-change. Finally, in light of the derivation, we elucidate the structure of mechanics by classifying the principles and assumptions that have been employed according to their explanatory role, distinguishing between symmetry principles and other types of principles (such as compositional principles) that are needed to build up the theoretical edifice.

Solution of Functional Equations

In this appendix, the functional equations needed in the derivation of the energy and momentum of bodies are solved. If possible, we transform the functional equation of interest into a standard functional equation. For interest, we sometimes provide more than one possible method of solution. In each case, certain mathematical conditions must be satisfied by the unknown function in order for a solution to be obtained.

1.1 Solution of $f(u + v) + f(u - v) = 2f(\sqrt{u^2 + v^2})$

We present two different solution methods for Eq. (1), one that transforms it into Jensen’s functional equation, the other a direct solution by removing one degree of freedom.

1.1.1 Solution by Transformation into Jensen’s Functional Equation

Using the substitution $k(w^2) = f(w)$, Eq. (1) becomes

$$\begin{aligned} k\left( [u^2 + v^2] + 2uv\right) + k\left( [u^2 + v^2] - 2uv\right) = 2k(u^2 + v^2). \end{aligned}$$

(A1)

Setting $x = u^2 + v^2$,  $y = 2uv$, we obtain

$$\begin{aligned} k(x + y) + k(x - y) = 2k(x), \end{aligned}$$

(A2)

which is Jensen’s equation, with x, y independently variable within $x>0, y>0$. If k is continuous, this equation, under the stated conditions, has general solution $k(z) = az + b$. As k is continuous whenever f is continuous,

$$\begin{aligned} f(v) = av^2 + b \end{aligned}$$

(A3)

is the general solution of Eq. (1) under the condition that f is continuous.

1.1.2 Direct Solution by Removal of One Degree of Freedom

Alternatively, one can directly solve Eq. (1) by removing one degree of freedom, albeit at the cost of the stronger regularity condition that f is analytic. Setting $u = v$ in Eq. (1) yields

$$\begin{aligned} f(2u) + f(0) = 2f(\sqrt{2}u). \end{aligned}$$

(A4)

If f is differentiable, then, for $n \ge 1$,

$$\begin{aligned} 2^n f^{(n)}(2u) = 2^{1 + n/2} \, f(\sqrt{2} u). \end{aligned}$$

(A5)

This yields $f^{(n)}(0) = 0$ whenever $n \ne 2$. Hence, if f is analytic,

$$\begin{aligned} f(x) = av^2 + b. \end{aligned}$$

(A6)

1.2 Solution of $g(v + u) - g(v - u) = 2g( \sqrt{u^2 + v^2}) \cdot v / \sqrt{u^2 + v^2}$

Solution of Eq. (6) is most readily obtained by removing one degree of freedom by setting $v = u$. Thence,

$$\begin{aligned} g(2u) - g(0) = \sqrt{2}\,g(\sqrt{2} u). \end{aligned}$$

(A7)

Setting $u = 0$ fixes $g(0) = 0$. If g is differentiable, then, for $n \ge 1$,

$$\begin{aligned} 2^n g^{(n)} (2u) = 2^{(n+1)/2} \, g^{(n)}(\sqrt{2} u). \end{aligned}$$

(A8)

For $n \ge 2$, this yields $g^{(n)}(0) = 0$. Thus, if g is analytic,

$$\begin{aligned} g(u) = au. \end{aligned}$$

(A9)

1.3 Solution of $\widetilde{F}(x) + \widetilde{F}(y) = 2\widetilde{F}((x+y)/2)$

Equation (34), with $x = \gamma (u \oplus -v)$ and $y = \gamma (u \oplus v)$, has the form of Jensen’s equation, but it is not immediately apparent that x, y are independent in some region. To see that this is so, it is helpful to express u, v in terms of rapidities:

$$\begin{aligned} \begin{aligned} u&= c \tanh \phi _1 \\ v&= c \tanh \phi _2. \end{aligned} \end{aligned}$$

(A10)

Then $u \oplus v = c\tanh (\phi _1 + \phi _2)$, so that

$$\begin{aligned} \begin{aligned} \gamma (u \oplus v)&= \tilde{\gamma }(\phi _1 + \phi _2) \\ \gamma (u \oplus -v)&= \tilde{\gamma }(\phi _1 - \phi _2), \end{aligned} \end{aligned}$$

(A11)

where $\tilde{\gamma }(\phi ) \equiv (1 - \tanh ^2\phi )^{-1/2}$.

Now, $u > 0$ and $|v| < c$, so that $\phi _1 > 0$ and $\phi _2$ is free. Consequently, $(\phi _1 + \phi _2)$ and $(\phi _1 - \phi _2)$ can be independently chosen. Further, since $\tilde{\gamma }$ is monotonic, $x = \tilde{\gamma }(\phi _1 + \phi _2)$ and $y = \tilde{\gamma }(\phi _1 - \phi _2)$ are independent in some region. Therefore, Eq. (34) has the solution $\widetilde{F}(x) = a + bx$.