A Cosmological PDE: Global Issues

In the previous post we addressed the solution of the following PDE

\left( \partial^2 - \frac{\kappa(\kappa+1)}{\eta^2} \right) \psi[\eta,\vec{x}] = J[\eta,\vec{x}]

by obtaining explicitly its associated Green’s function G[x,x']. In this post, we shall turn to discussing global issues. First, we will generalize the PDE to

D_x \psi[x] \equiv \left( \partial^2 - \frac{\ddot{a}[\eta]}{a[\eta]} \right) \psi[\eta,\vec{x}] = J[\eta,\vec{x}];

where the scale factor a[\eta] describes the relative size of the Universe; with the Big Bang occurring at a[\eta=0]=0; and the Universe subsequently expanding for \eta>0. Keeping in mind

D_{x} G[x,x'] = \delta[\eta-\eta'] \delta^{(3)}[\vec{x}-\vec{x}'] = D_{x'} G[x,x'];

let us therefore consider the following:

\psi[\eta>0,\vec{x}] = \int_0^\infty d\eta' \int_{\mathbb{R}^3} d^3 \vec{x}' \left( D_{x'} G[x,x'] \cdot \psi[x'] - G[x,x'] \cdot D_{x'} \psi[x'] \right) \\ \qquad \qquad + \int_0^\infty d\eta' \int_{\mathbb{R}^3} d^3 \vec{x}' G[x,x'] \cdot J[x'] .

Integration-by-parts and the assumption that \psi at spatial infinity is zero yield

\psi[\eta>0,\vec{x}] = \int_{\mathbb{R}^3} d^3 \vec{x}' \left[ \partial_{\eta'} G[x,x'] \cdot \psi[x'] - G[x,x'] \cdot \partial_{\eta'} \psi[x'] \right]_{\eta'=0^+}^{\eta'=+\infty} \\ \qquad \qquad + \int_0^\infty d\eta' \int_{\mathbb{R}^3} d^3 \vec{x}' G[x,x'] \cdot J[x'] .

If we further employ the retarded Green’s function, so that G[\eta < \eta' \to +\infty] = 0,

(1) \psi[\eta>0,\vec{x}] = \int_{\mathbb{R}^3} d^3 \vec{x}' \left( G[x,x'] \cdot \partial_{\eta'} \psi[x'] - \partial_{\eta'} G[x,x'] \cdot \psi[x'] \right)_{\eta'=0^+} \\ \qquad \qquad + \int_0^\infty d\eta' \int_{\mathbb{R}^3} d^3 \vec{x}' G[x,x'] \cdot J[x'] .

If \eta had ran over \mathbb{R} the first term involving a spatial volume integral at the Big Bang would not be present. Furthermore, on physical grounds, is it not strange that the field and its velocity needs to be specified at the Big Bang — where geometric tidal forces are infinite — for the field to be properly determined for \eta > 0?

Gauss’ Law Violation

What sort of trouble would arise if we had simply dropped the Big Bang initial conditions spatial integral in eq. (1)? Firstly, let us notice that, the integral of the Green’s function against the source yields the concept of the particle horizon — namely, the furthest portion of the signal from any physical and, hence, timelike source J at a given time is simply the wavefront emanating from the latter at the Big Bang. Any object lying beyond the particle horizon cannot be influenced by J because the integral involving it in eq. (1) is zero there.

The equations of electromagnetic and linearized gravitational fields imply a global Gauss’ law type of constraint, that the flux of the electric field (and its analogous gravitational one) across some closed 2D spatial surface at a given time, must in fact be equal to the total electric charge (or, total gravitational mass) contained within this surface. However, the particle horizon poses the following puzzle. If we had dropped the initial conditions term in eq. (1), so that the relevant electromagnetic or linearized gravitational fields due to some source J is strictly zero outside its particle horizon, how is Gauss’ law obeyed if this 2D closed surface lies outside of it? In fact, this is why the initial conditions become critical for global consistency of the solutions: they must be correctly specified so that the electromagnetic and gravitational field outside the particle horizon are entirely given by the initial data in such a manner consistent with Gauss’ law.

Author: Yi-Zen Chu

I am a theoretical physicist, with research interests spanning gravitation and field theory, particle cosmology and Mathematica software development.

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