Dynamics · Force and Work

01Premises and reference parameter · Starting Assumptions

Under these conditions $dW=\mathbf v\cdot d\mathbf p$. Internal excitation, radiation or a changing background requires its own momentum flux and energy exchange rather than only the fixed-$m_0$ translational term.

02Differentiating total momentum · Momentum Derivative

Apply the product rule to $\mathbf p=m_0\gamma\mathbf v$:

Apply the chain rule to $\gamma=(1-\mathbf v\cdot\mathbf v/c^2)^{-1/2}$:

This differentiates the scale factor along the motion process. No time-dilation premise is inserted alongside the product rule.

03The response by direction · Directional Response

Substitute $\dot\gamma$ into the force expression:

Resolve $\mathbf a=\mathbf a_\parallel+\mathbf a_\perp$, with $\mathbf v\cdot\mathbf a_\perp=0$. The coefficient along velocity is

whereas the perpendicular coefficient is $\gamma$. Both coefficients arise from the derivative of one momentum with respect to velocity, not from two rest masses. At $v=0$, the directional decomposition is arbitrary and the total response approaches $m_0\mathbf a$ continuously.

Inverting the force response

The longitudinal and transverse eigenvalues of the response tensor are $m_0\gamma^3$ and $m_0\gamma$. Both are positive for $m_0>0$ and $v<c$, so the force law can be inverted without a low-speed expansion:

This is final-manuscript equation (2.20). It gives the acceleration produced by a specified total external force, while (2.18) gives the force required for a specified acceleration. At zero speed it continuously becomes $\mathbf a=\mathbf F/m_0$. The separately retained low-speed Newtonian expansion is a website check of (2.18), not the content of final equation (2.20).

04Longitudinal, transverse and circular motion · Special Cases

For uniform circular motion in one local calibration, $a_\perp=v^2/R$, so

Here $v$ and $R$ use the same local length and counting reference. Since $\gamma$ already enters momentum, replacing $R$ by $\gamma R$ without an additional measurement map would count a scale effect again. Comparing a different length realization requires that separate map first. Chapter 2 explicitly preserves this condition after Eq. (2.21).

05From work to kinetic energy · Work and Energy

When only speed changes, $dp/dv=m_0\gamma^3$, allowing integration from rest to $v$. A change of momentum perpendicular to velocity contributes zero to the dot product even when the direction changes.

See Energy and Action for the complete integral and total-energy normalization. A work-derived energy difference does not already determine rest energy.

06Physical meaning and domain · Interpretation and Validity

ASD interprets the directional force response as reconfiguration of a moving spatial structure. Longitudinal acceleration changes speed and mean gain; transverse acceleration changes direction. The instantaneous force in uniform circular motion is perpendicular to velocity and does not change kinetic energy in this model. If a charged structure radiates, that radiation must be added.

The expressions apply at fixed mass, in a uniform background, with differentiable velocity and $v<c$. External gravity, material stress and many-body processes do not acquire a complete dynamics merely from matching free-motion formulas.

07Consistency and sources · Checks and Sources

Dimensions are mass times acceleration. The $\beta\to0$ limit recovers Newtonian response, and $\mathbf v\cdot\mathbf a=0$ gives $\dot\gamma=0$. Power $\mathbf F\cdot\mathbf v$ equals the process derivative of kinetic energy. See Experimental Relations for mathematical agreement and interpretive differences.

This page preserves Eqs. (2.15)–(2.21) and the unnumbered circular-motion relation. Intermediate differentiations expand the manuscript’s algebra without adding Lorentz transformations as premises.