Below is a clean, step-by-step derivation showing how the term $L^{2}/r^{3}$ arises when going from the EIH equations of motion to the radial equation for a two-body system. I will keep the discussion strictly analytical and structural, not heuristic.

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1. Start point: EIH equation (two-body, weak field, slow motion)

For two masses $m_1, m_2$, the Einstein–Infeld–Hoffmann (EIH) equations give, to first post-Newtonian (1PN) order, the acceleration of body 1:

$$\ddot{\mathbf r} = - \frac{G M}{r^3}\mathbf r + \frac{1}{c^2}\mathbf a_{\text{1PN}}$$

where

• $M = m_1 + m_2$

• $\mathbf r = \mathbf r_1 - \mathbf r_2$

• $r = |\mathbf r|$

The Newtonian term is already explicit:

$$- \frac{GM}{r^3}\mathbf r$$

The 1PN correction contains velocity-dependent and potential-dependent terms but does not create angular momentum by itself. Angular momentum arises from rotational symmetry.

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2. Reduce to relative motion (center-of-mass frame)

Introduce reduced mass

$$\mu = \frac{m_1 m_2}{m_1 + m_2}$$

The EIH equations admit a Lagrangian formulation. To Newtonian order:

$$L = \frac{1}{2}\mu \dot{\mathbf r}^2 + \frac{GM\mu}{r}$$

The existence of a Lagrangian is crucial: it implies conserved quantities.

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3. Rotational invariance → conserved angular momentum

Because the Lagrangian depends only on $r$, not on orientation, angular momentum is conserved:

$$\mathbf L = \mu \mathbf r \times \dot{\mathbf r}$$

Its magnitude is constant:

$$L = \mu r^2 \dot{\phi}$$

This relation is exact at Newtonian order and remains valid at 1PN order up to small corrections that do not change the structure of the $L^2/r^3$ term.

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4. Decompose motion into radial + angular parts

Write velocity squared:

$$\dot{\mathbf r}^2 = \dot r^2 + r^2 \dot{\phi}^2$$

Insert angular momentum:

$$r^2 \dot{\phi}^2 = \frac{L^2}{\mu^2 r^2}$$

So the kinetic term becomes:

$$\frac{1}{2}\mu \dot{\mathbf r}^2 = \frac{1}{2}\mu \dot r^2 + \frac{L^2}{2\mu r^2}$$

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5. Construct the effective radial Lagrangian

The radial Lagrangian is:

$$L_{\text{radial}} = \frac{1}{2}\mu \dot r^2 - \left( \frac{L^2}{2\mu r^2} - \frac{GM\mu}{r} \right)$$

Define the effective potential:

$$V_{\text{eff}}(r) = \frac{L^2}{2\mu r^2} - \frac{GM\mu}{r}$$

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6. Radial equation of motion

Apply Euler–Lagrange:

$$\mu \ddot r = -\frac{dV_{\text{eff}}}{dr}$$

Differentiate:

$$\frac{d}{dr}\left(\frac{L^2}{2\mu r^2}\right) = - \frac{L^2}{\mu r^3}$$ $$\frac{d}{dr}\left(-\frac{GM\mu}{r}\right) = \frac{GM\mu}{r^2}$$

Therefore:

$$\mu \ddot r = \frac{L^2}{\mu r^3} - \frac{GM\mu}{r^2}$$

Divide by $\mu$:

$$\boxed{ \ddot r = \frac{L^2}{\mu^2 r^3} - \frac{GM}{r^2} }$$

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7. Origin of the $L^2/r^3$ term — precise statement

The term $L^2/r^3$ is NOT introduced by gravity.

It arises from:

1. Rotational symmetry of the EIH Lagrangian

2. Conservation of angular momentum

3. Elimination of angular degrees of freedom

4. Differentiation of the centrifugal potential $L^2/(2\mu r^2)$

This is a purely kinematic consequence, already present in Newtonian mechanics, and remains structurally unchanged in the EIH framework.

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8. Relation to full EIH / 1PN corrections

At 1PN order, the effective potential gains extra terms:

$$V_{\text{eff}}(r) = \frac{L^2}{2\mu r^2} - \frac{GM\mu}{r} - \frac{GML^2}{c^2 \mu r^3} + \cdots$$

Note carefully:

• The classical $L^2/r^3$ term comes from geometry

• The relativistic correction $GM L^2 / (c^2 r^3)$ comes from spacetime curvature

This distinction is essential.

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9. Final summary (one-line)

$$\boxed{ \text{The } \frac{L^2}{r^3} \text{ term arises from angular momentum conservation after eliminating angular motion, not from the EIH force itself.} }$$

If you want, I can next:

• Show how the additional $1/r^3$ term from GR differs conceptually

• Derive the perihelion precession directly from EIH

• Map this explicitly onto your second- vs third-order space-time ontology

Just indicate which direction you want to go.