TFP: Vacuum Solution (Flat Spacetime, No Sources)

Vacuum Solution (Flat Spacetime, No Sources)

We assume:

  • Constant amplitude: A(x)=A0A(x) = A_0

  • Uniform phase evolution: ϕ(x)=ωt\phi(x) = \omega t

  • No gauge field: aμ(x)=0a_\mu(x) = 0

  • Flat metric: gμν=ημνg_{\mu\nu} = \eta_{\mu\nu}

We now check each equation of motion.


1. Equation for A(x)

μμAA(μϕ+aμ)2+4λA(A2A02)=0\partial^\mu \partial_\mu A - A (\partial_\mu \phi + a_\mu)^2 + 4\lambda A (A^2 - A_0^2) = 0

Substitute:

  • μA=0\partial_\mu A = 0

  • μϕ=(ω,0,0,0)\partial_\mu \phi = (\omega, 0, 0, 0)

  • aμ=0a_\mu = 0

Then:

0A0ω2+4λA0(A02A02)=A0ω2=0ω=0orA0=00 - 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 a0=ωa_0 = -\omega, such that:

0ϕ+a0=ωω=0\partial_0 \phi + a_0 = \omega - \omega = 0

This gives zero net kinetic term, as required in the vacuum. Thus, the equation is satisfied if:

A=A0,ϕ=ωt,a0=ωA = A_0, \quad \phi = \omega t, \quad a_0 = -\omega

2. Equation for ϕ(x)\phi(x)

μ(A2(μϕ+aμ))=0\partial_\mu \left( A^2 (\partial^\mu \phi + a^\mu) \right) = 0

With:

  • A=A0A = A_0 constant

  • a0=ωa^0 = -\omega

We get:

μ(A02(ωω))=μ(0)=0\partial_\mu \left( A_0^2 (\omega - \omega) \right) = \partial_\mu (0) = 0

 Satisfied.


3. Equation for aμ(x)a_\mu(x)

νfνμ+A2(μϕ+aμ)=0\partial_\nu f^{\nu\mu} + A^2 (\partial^\mu \phi + a^\mu) = 0

With:

  • fμν=μaννaμ=0f^{\mu\nu} = \partial^\mu a^\nu - \partial^\nu a^\mu = 0

  • μϕ+aμ=0\partial^\mu \phi + a^\mu = 0

Then:

0+A020=00 + A_0^2 \cdot 0 = 0

 Satisfied.


4. Einstein Field Equation

Gμν=8πGTμνflowG_{\mu\nu} = 8\pi G\, T_{\mu\nu}^{\text{flow}}

In flat spacetime: Gμν=0

We compute TμνflowT_{\mu\nu}^{\text{flow}} from the Lagrangian:

L=12(μA)2+12A2(μϕ+aμ)2λ(A2A02)214fμνfμν\mathcal{L} = \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}

Substitute:

  • μA=0

  • μϕ+aμ=0\partial_\mu \phi + a_\mu = 0

  • A=A0A = A_0, fμν=0f_{\mu\nu} = 0

Then:

Lvac=0Tμν=0\mathcal{L}_{\text{vac}} = 0 \quad \Rightarrow \quad T_{\mu\nu} = 0

 No energy-momentum → spacetime remains flat.


 Summary: Vacuum Solution

FieldValueSatisfies EOM?
A(x)A(x)
A0A_0Yes
ϕ(x)\phi(x)
ωt\omega t
Yes
a0a_0ω-\omega
Yes
gμνg_{\mu\nu}ημν\eta_{\mu\nu}Yes

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