Dist Babble

Perturbations Around the Vacuum Start from the vacuum solution: A ( x ) = A 0 A(x) = A_0 ​  (constant amplitude) ϕ ( x ) = ω t \phi(x) = \omega t  (background phase) a μ ( x ) = − ω δ μ 0 a_\mu(x) = -\omega \delta^0_\mu  (cancels time evolution) g μ ν = η μ ν g_{\mu\nu} = \eta_{\mu\nu}  (flat spacetime) Now introduce small perturbations : ϕ ( x ) = ω t + δ ϕ ( x ) \phi(x) = \omega t + \delta\phi(x) a μ ( x ) = − ω δ μ 0 + δ a μ ( x ) a_\mu(x) = -\omega \delta^0_\mu + \delta a_\mu(x) These combine in the covariant derivative as: D μ ϕ = ∂ μ ϕ + a μ = δ μ 0 ω + ∂ μ δ ϕ + δ a μ − ω δ μ 0 = ∂ μ δ ϕ + δ a μ D_\mu \phi = \partial_\mu \phi + a_\mu = \delta_\mu^0 \omega + \partial_\mu \delta\phi + \delta a_\mu - \omega \delta_\mu^0 = \partial_\mu \delta\phi + \delta a_\mu ​ So the entire dynamics will be governed by the perturbations: χ μ ( x ) ≡ ∂ μ δ ϕ + δ a μ \chi_\mu(x) \equiv \partial_\mu \delta\phi + \delta a_\mu ​ Equation of Motion for δ a μ \delta a_\mu ...

Vacuum Solution (Flat Spacetime, No Sources) We assume: Constant amplitude:  A ( x ) = A 0 A(x) = A_0 ​ Uniform phase evolution:  ϕ ( x ) = ω t \phi(x) = \omega t No gauge field:  a μ ( x ) = 0 a_\mu(x) = 0 Flat metric:  g μ ν = η μ ν g_{\mu\nu} = \eta_{\mu\nu} ​ We now check each equation of motion. 1. Equation for A ( x ) ∂ μ ∂ μ A − A ( ∂ μ ϕ + a μ ) 2 + 4 λ A ( A 2 − A 0 2 ) = 0 \partial^\mu \partial_\mu A - A (\partial_\mu \phi + a_\mu)^2 + 4\lambda A (A^2 - A_0^2) = 0 Substitute: ∂ μ A = 0 \partial_\mu A = 0 ∂ μ ϕ = ( ω , 0 , 0 , 0 ) \partial_\mu \phi = (\omega, 0, 0, 0) a μ = 0 a_\mu = 0 Then: 0 − A 0 ω 2 + 4 λ A 0 ( A 0 2 − A 0 2 ) = − A 0 ω 2 = 0 ⇒ ω = 0 or A 0 = 0 0 - A_0 \omega^2 + 4\lambda A_0 (A_0^2 - A_0^2) = -A_0 \omega^2 = 0 \quad \Rightarrow \quad \omega = 0 \quad \text{or} \quad A_0 = 0 But we want non-zero flow. So to satisfy the equation, we must reintroduce a 0 = − ω a_0 = -\omega , such that: ∂ 0 ϕ + a 0 = ω − ω = 0 \partial_0 \ph...

  Equation of Motion for A ( x ) A(x)  (Flow Amplitude) We start with the full action: S = ∫ d 4 x [ 1 2 ( ∂ μ A ) 2 + 1 2 A 2 ( ∂ μ ϕ + a μ ) 2 − λ ( A 2 − A 0 2 ) 2 − 1 4 f μ ν f μ ν ] S = \int d^4x \left[ \frac{1}{2} (\partial_\mu A)^2 + \frac{1}{2} A^2 (\partial_\mu \phi + a_\mu)^2 - \lambda (A^2 - A_0^2)^2 - \frac{1}{4} f_{\mu\nu} f^{\mu\nu} \right] Take variation with respect to A ( x ) A(x) : δ S = ∫ d 4 x [ δ A ( − ∂ μ ∂ μ A + A ( ∂ μ ϕ + a μ ) 2 − 4 λ A ( A 2 − A 0 2 ) ) ] \delta S = \int d^4x \left[ \delta A \left( -\partial^\mu \partial_\mu A + A (\partial_\mu \phi + a_\mu)^2 - 4\lambda A (A^2 - A_0^2) \right) \right] Set δ S = 0 \delta S = 0  ⇒ Euler–Lagrange equation for A ( x ) A(x) : ∂ μ ∂ μ A − A ( ∂ μ ϕ + a μ ) 2 + 4 λ A ( A 2 − A 0 2 ) = 0 \boxed{ \partial^\mu \partial_\mu A - A (\partial_\mu \phi + a_\mu)^2 + 4\lambda A (A^2 - A_0^2) = 0 } ​ This governs how the magnitude of temporal flow evolves — driven by phase velocity ∂ μ ϕ \partial_\mu \phi...