TFP: Motion in the static, spherically symmetric case

Motion in the static, spherically symmetric case

We assume:

• Static central source, at spatial origin.

• Time-independent fields:
  $A(x,t) \to A(r)$, $\phi(x,t) \to \phi(r)$, $a_\mu(x,t) \to a_0(r), a_i = 0$
  

• Spherical symmetry: all fields depend only on the radial coordinate $r$.

• Weak field metric (for now):
  $ds^2 = - (1 + 2\Phi(r))\,dt^2 + (1 - 2\Psi(r))\,dr^2 + r^2 d\Omega^2$
  

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Step 2: Equations of Motion Recap (Simplified for Static Case)

From earlier, the coupled Lagrangian density was:

$$\mathcal{L} = \frac{1}{2} (\nabla A)^2 + \frac{1}{2} A^2 (\nabla \phi)^2 - V(A) - \frac{1}{4} f_{\mu\nu}f^{\mu\nu} + \lambda A a^\mu \partial_\mu \phi + \text{(gravity terms)}$$

We now extract the Euler-Lagrange equations in this static case:

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1. Equation for $A(r)$:

$$\nabla^2 A - A (\nabla \phi)^2 - \frac{dV}{dA} + \lambda a^r \partial_r \phi = 0$$

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2. Equation for $\phi(r)$:

$$\nabla \cdot (A^2 \nabla \phi) + \lambda \nabla \cdot (A a^r) = 0$$

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3. Equation for $a^0(r)$ (only non-zero component):

Maxwell-like equation:

$$\nabla^2 a^0(r) = \lambda A(r) \frac{d\phi}{dr}$$

This implies the charge density is sourced by the temporal flow phase gradient weighted by A, as previously discussed.

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4. Einstein equation for $g_{\mu\nu}$ (static, weak field):

We focus on the Newtonian limit $G_{00} \sim -2\nabla^2 \Phi(r)$:

$$\nabla^2 \Phi(r) = 4\pi G \rho_{\text{eff}}(r)$$

Where $\rho_{\text{eff}}$ comes from the energy density in $A, \phi, a^0$, and their interactions.

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Step 3: Solve in Stages

Let's work through the simplest case first — a static point source, where:

• $V(A) = \frac{1}{2} m_A^2 (A - A_0)^2$
  

• $\phi'(r) = \frac{q}{4\pi r^2 A(r)}$(from Gauss's law-like structure)

• Central charge density gives $\nabla^2 a^0(r) = \delta^3(r)$ (or smoothed version)

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🔹 Solve Maxwell-like Equation

From:

$$\nabla^2 a^0 = \lambda A(r) \phi'(r)$$

Insert the above estimate for $\phi'(r)$:

$$\nabla^2 a^0 = \lambda A(r) \cdot \frac{q}{4\pi r^2 A(r)} = \frac{\lambda q}{4\pi r^2} \Rightarrow \nabla^2 a^0 = \frac{\lambda q}{4\pi r^2}$$

This integrates to:

$$a^0(r) = -\frac{\lambda q}{4\pi r} + \text{const}$$

✅ Matches Coulomb potential!

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🔹 Solve for $\phi(r)$

From Gauss-like law:

$$\nabla \cdot (A^2 \nabla \phi) = -\lambda \nabla \cdot (A a^r)$$

If $a^r = 0$, then:

$$\frac{1}{r^2} \frac{d}{dr} \left( r^2 A^2 \frac{d\phi}{dr} \right) = 0 \Rightarrow A^2 \frac{d\phi}{dr} = \frac{\text{const}}{r^2} \Rightarrow \frac{d\phi}{dr} = \frac{q}{4\pi r^2 A^2}$$

So $\phi(r)$ is again logarithmic or Coulomb-like, depending on A(r).

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🔹 Solve for $A(r)$

Now plug the expression for $\phi'(r)$ into the equation for A:

$$\nabla^2 A - A \left( \frac{q^2}{(4\pi)^2 r^4 A^4} \right) - m_A^2 (A - A_0) = 0$$

This is a nonlinear second-order ODE for A(r). But in the weak field limit, we linearize:

Let $A(r) = A_0 + \delta A(r)$, then the equation becomes:

$$\nabla^2 \delta A - m_A^2 \delta A - \frac{q^2}{(4\pi)^2 r^4 A_0^3} = 0$$

Which is solvable — the solution will show screening (Yukawa-type decay) if $m_A \neq 0$, and a slower decay if massless.

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🔹 Solve for Gravitational Potential $\Phi(r)$

We use the stress-energy from $A, \phi, a^0$, compute the energy density, and plug into:

$$\nabla^2 \Phi(r) = 4\pi G \rho_{\text{eff}}(r)$$

The dominant contribution (for static point charge) will yield:

$$\Phi(r) \sim -\frac{G M_{\text{eff}}}{r}$$

Where $M_{\text{eff}}$ includes mass from A-field condensation, energy in $\phi$ gradients, and possibly in $a^0$.

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✅ Summary of Results

Quantity	Result
$a^0(r)$	$-\dfrac{\lambda q}{4\pi r}$ → Coulomb potential
$\phi(r)$	$\int \dfrac{q}{4\pi r^2 A(r)^2} dr$ → phase gradient structure
$A(r)$	Solves nonlinear ODE; approximates vacuum + Yukawa screening
$\Phi(r)$	Newtonian potential with effective mass from flow energy