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    "Premises and energy reference · Starting Assumptions",
    "From momentum to work · Work Integral",
    "Essential integration steps · Intermediate Steps",
    "Rest energy and total energy · Rest and Total Energy",
    "Energy–momentum consistency · Algebraic Consistency",
    "Free-motion action · Free Lagrangian",
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    "Sources and experimental interface · Sources"
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---
Work done by external forces enters the joint motion reconstruction of the core and accompanying field. Integrating $dE_k=\mathbf v\cdot d\mathbf p$ gives kinetic energy; adding the stored energy of the same rest structure yields Chapter 2’s continuous energy–momentum construction.

## Premises and energy reference · Starting Assumptions

<AssumptionCard id="A1" title="Retain the derived momentum and work definition">

Use $\mathbf p=m_0\gamma\mathbf v$ and $dE_k=\mathbf v\cdot d\mathbf p$. The background is uniform, internal state fixed, and net radiative loss absent. Set kinetic energy to zero at $v=0$. Source: Eqs. (2.14) and (2.22).

</AssumptionCard>

<AssumptionCard id="A2" title="Declare the rest-energy normalization separately">

For total energy, Chapter 2 adopts $E_0=m_0c^2$. This is not needed to integrate the kinetic-energy difference. It is a normalization input linking static structural energy to the inertia parameter. Source: text between Eqs. (2.23) and (2.24).

</AssumptionCard>

## From momentum to work · Work Integral

Differentiate $p=m_0\gamma v$:

$$
\frac{dp}{dv}=m_0\left(\gamma+\gamma^3\frac{v^2}{c^2}\right)=m_0\gamma^3.
$$

Kinetic energy from rest to speed $v$ is therefore:

<Equation name="dyn-work-integral" />

$w$ is the integration speed. For a general change of direction, $\mathbf v\cdot d\mathbf p$ retains only the speed-changing component. Under the stated conditions, kinetic energy is a function of speed.

## Essential integration steps · Intermediate Steps

Set $z=1-w^2/c^2$, giving $w\,dw=-c^2dz/2$. The lower limit $w=0$ gives $z=1$, and the upper limit $w=v$ gives $z=1-v^2/c^2$:

$$
\begin{aligned}
E_k&=-\frac{m_0c^2}{2}\int_1^{1-v^2/c^2}z^{-3/2}\,dz\\
&=-\frac{m_0c^2}{2}\left[-2z^{-1/2}\right]_1^{1-v^2/c^2}\\
&=m_0c^2\left[\left(1-\frac{v^2}{c^2}\right)^{-1/2}-1\right].
\end{aligned}
$$

This is the unnumbered calculation following manuscript Eq. (2.22); it does not work backwards from a standard energy formula.

<Equation name="dyn-kinetic-energy" />

Using $\gamma=1+\beta^2/2+3\beta^4/8+O(\beta^6)$ recovers Newtonian kinetic energy and its next correction. Both coefficients come from the expansion of one function.

## Rest energy and total energy · Rest and Total Energy

<ArgumentStep kind="Normalization" title="Static structural energy requires independent matching">

ASD treats a structure at rest as carrying strain and structural energy. To connect this energy to its inertia parameter, the chapter adopts:

<Equation name="dyn-rest-energy" />

</ArgumentStep>

The work integral determines energy differences and cannot fix $E_0$ by itself. With this normalization:

<Equation name="dyn-total-energy" />

Thus $E=m_0\gamma c^2$ combines the kinetic-energy result and A2. A microscopic theory must compute $E_0$ and $m_0$ from the same stable structural solution to test the normalization; it is not a fact established by this integral alone.

## Energy–momentum consistency · Algebraic Consistency

$$
E^2-c^2p^2
=m_0^2c^4\gamma^2\left(1-\frac{v^2}{c^2}\right)
=m_0^2c^4.
$$

<Equation name="dyn-energy-identity" />

This is an algebraic consistency relation after eliminating $v$. In this chapter it follows $p$ and $E$; a presumed Minkowski invariant does not replace ASD’s constitutive derivation.

## Free-motion action · Free Lagrangian

Require the free translational Lagrangian to satisfy $\partial\mathcal L_0/\partial\mathbf v=\mathbf p$. Integrate over speed:

$$
\mathcal L_0(v)-\mathcal L_0(0)
=\int_0^v\frac{m_0w}{\sqrt{1-w^2/c^2}}\,dw
=m_0c^2\left(1-\sqrt{1-v^2/c^2}\right).
$$

Fix the constant with $\mathcal L_0(0)=-E_0=-m_0c^2$, giving:

<Equation name="dyn-lagrangian" />

Direct differentiation recovers $m_0\gamma\mathbf v$. The Legendre transform gives

$$
\mathbf p\cdot\mathbf v-\mathcal L_0
=m_0c^2\left(\gamma\beta^2+\frac1\gamma\right)
=m_0c^2\gamma.
$$

The website uses $\mathcal L_0$ to distinguish the Lagrangian from length $L$; it corresponds to $L_0$ in manuscript Eq. (2.24b). Its low-speed expansion is

$$
\mathcal L_0=-m_0c^2+\frac{m_0v^2}{2}+\frac{m_0v^4}{8c^2}+O\!\left(\frac{m_0v^6}{c^4}\right).
$$

Chapter 2’s later orbital calculation must retain the $v^4$ term alongside gravitational terms of the same order. This module supplies that interface without developing the orbital approximation here.

## Physical meaning, scope and review · Interpretation and Review

Kinetic energy describes reconfiguration of the moving structure; rest energy describes the static core and accompanying structure. Matching mathematical forms do not remove these distinct physical roles.

For $0\leq v<c$ and positive fixed $m_0$, $E_k\geq0$, $E(0)=E_0$ and $dE_k/dt=\mathbf F\cdot\mathbf v$. Divergence as $v\to c$ is a feature within this model, not an account of crossing that boundary. Radiation, internal excitation and background changes require an expanded energy balance.

Human review concerns whether inertia closure and static energy derived from one microscopic structure support the common normalization. That question is distinct from the correctness of the definite integral; see [OP-DYN-01 (Chinese)](/open-problems/inertial-closure).

## Sources and experimental interface · Sources

This page follows Chapter 2 §III, final manuscript §III, Eqs. (2.22)–(2.24b) and the unnumbered intermediate steps. Universal coupling of energy, inertial and gravitational mass is discussed in [Experimental Relations](/en/theory/dynamics/tests).

<CanonicalSource />
