# Infinite distribution of charge vs infinite distribution of matter

Physics Asked on May 9, 2021

From my textbook:

General Relativity is the currently most complete theory of Gravitation,
and should be used to describe physical systems each time one of the
following assumptions does not hold:

• Gravitational fields are small, $$Phi << c^2$$
• The scale of the system is much smaller than the curvature radius
• Velocities are small $$v << c$$

A way to satisfy these hypotheses in cosmology is to consider the evolution
of a small spherical portion of the Universe, where Newton’s theory
applies: then, the behaviour of this portion will reflect the evolution of
the Universe (see below).
Two important notes:

• We can use the Newtonian Gravitation here only because we can
neglect the matter outside the sphere does not affect the field gravitational
field inside the sphere: this is a consequence of Gauss’s
theorem, which in GR is known as Birchoff’s theorem.

Important part:

In principle, the same applies to an infinite distribution of charges.
However there is a fundamental difference between the electrical
and the gravitational field: the former is measurable, while the second
is not (because of the equivalence principle). In gravitation,
only tidal fields can be measured! As a result, an infinite distribution
of charges does not make sense (for a finite sphere of charges,
the electrical field gives the center of the sphere, so the limit $$Rto infty$$
is not allowed: where is the center then?). But an infinite distribution
of matter is allowed in Newton’s theory (the gravitational field
increases linearly with R, which means that the tidal field is constant;
also, the tidal fields is isotropic and no direction for a center
is provided!)

• Why is the tidal field constant? I don’t understand this.
• Shouldn’t the gravitational field decrease linearly with $$R$$? (there is written increase).
• So the main difference is that the Electric field goes as $$displaystylefrac{1}{R^2}$$ outside a charged sphere while the Gravitational Field goes as $$displaystylefrac{1}{R}$$, right?
• Is really the Birchoff theorem the GR version of the Gauss
theorem?

We're interested in what happens inside the sphere, not outside it, since the sphere will fill all of space as $$Rrightarrow infty$$. Inside the sphere, both the electric field and the gravitational field are proportional to $$R$$. "Increase" is correct.

The main difference is not that the electric field and the gravitational field behave differently. The main difference is that the gravitational field is not observable. The tidal field is the observable thing.

Answered by Daniel on May 9, 2021