# What is the gravitational field intensity of a uniformly distributed mass content in Newtonian gravity?

In an infinite universe composed of single point masses which can be simplified as a uniformly distributed mass density, what is the equation for the gravitational field intensity in Newtonian gravity?

I assume that:

1. The gravitational field intensity is constant through space since the matter density is constant.

2. The net force on any test particle must be zero, because all forces must cancel out due to symmetry.

3. The gravitational field intensity is infinite since the its the sum the gravitational field intensity of each mass point, and there are endless mass points at infinite distance. But it must converge to a specific value since contributions of far away point masses contribute less and less with the square of distance.

• What do you mean by a uniformly distributed mass densities, as in the entire universe is filled with say....dust or small point masses but at every point in space? Commented Aug 16, 2023 at 3:46
• @Triatticus Yes. Commented Aug 16, 2023 at 3:49
• If you're familiar with newtons shell theorem, then you've already got the answer, it's more interesting of course if there are slight disturbances spoiling the symmetry though I know that's not necessarily what you're asking about. Commented Aug 16, 2023 at 3:56
• That doesn't matter here, an infinite plate of mass has the same field structure for gravity as a charged plate does for electric fields. You can show this as both integrals are identical to one another. Commented Aug 16, 2023 at 4:25
• Related question here. Commented Aug 19, 2023 at 22:02

In the frame of an inertial observer at some position, which we can take to be $$\vec r=0$$, the Newtonian gravitational field at position $$\vec r$$ is

$$\vec g\equiv \ddot{\vec r} = -\frac{4\pi G}{3}\rho \,\vec r,\tag{1}$$

where $$\rho$$ is the density of the universe. This follows straightforwardly from the shell theorem. Of course, it's also just the second Friedmann equation for the case of a universe containing only nonrelativistic matter.

That seems too easy! What about the homogeneity of the universe, which equation (1) does not seem to respect? Well, there are two important points to be made.

#### Different parts of the universe are accelerating with respect to each other.

This is due to their mutual gravitational attraction and is precisely in accordance with equation (1). A static universe is inconsistent with Newtonian gravity.

Indeed, while it's often suggested that cosmic expansion is a general relativistic effect, it's also predicted by Newtonian gravity.

#### The gravitational field is frame dependent.

Specifically, it depends on the reference frame's acceleration. Since different parts of the universe are accelerating with respect to each other, their inertial frames (that is, frames only accelerated by gravity) measure different gravitational fields. In particular, the local gravitational field is always zero for an inertial observer. As a function of position, it follows equation (1) for every observer (if they define their own position to be $$\vec r = 0$$).

This is how equation (1) respects the symmetry of a uniform mass distribution.

1. For any sphere with uniform density, it can be shown by integration, that the acceleration of gravity increases linearly with the distance to the centre. And it is zero at the centre. It is supposed no mass outside it.

2. It can also be shown that the acceleration of gravity inside a spherical shell is zero.

Both results implies that inside a sphere the conclusion of (1) is valid, even with masses out of the sphere, provided that there is a spherical symmetry outside.

But for an infinite sphere the centre can be anywhere. And if the acceleration of gravity is zero at the centre of the sphere, we can say that there is no acceleration at any point.

But on the other hand, if a given point is selected to be the centre, we can also say that the nearby points are off-centre, what leads to different conclusion.

So, I don't see how to model the behaviour of an infinite universe using Newtonian gravity.

We can also use the field equation for Newtonian gravity: $$\nabla^2 \phi = -4\pi G \rho$$ for $$\rho =$$ constant.

But the general solution: $$\phi = -G \rho \int \frac{dr'^3}{|r-r'|}$$ diverges when integrating from zero to infinity.

• The acceleration must be zero because all gravitational forces of a test particle placed anywhere must cancel out (the exterior is symmetrical). That does not mean the gravitational field intensity is zero. I want to know the value for the gravitational field intensity. Commented Aug 19, 2023 at 20:27