From all that I have heard about Newton's Law of Universal Gravity, one fact, which I find quite interesting, is that the distance between the two objects of mass is squared and not cubed due to our Universe only having three spacial dimensions. By this logic, if we lower the amount of dimensions to two dimensions, then Newtonian gravity in a 2D space should follow a regular inverse law with no squaring, due to having the distance raised to the first power.

Now, when we have only one dimension of space, the distance ends up being raised to the power of 0, meanining it is 1, so the distance between the two objects of mass becomes seemingly irrelevant when measuring the strength of gravity in 1D space. I want to understand whether it is just due to the particular formula breaking down or would distance really not matter for strength of gravity in one-dimensional space?

  • $\begingroup$ How does mass scale with reduction in dimension in your model? As you know, Newton's law depends also on the masses of the objects in question. Thanks. $\endgroup$
    – ad2004
    Commented May 25 at 3:51

2 Answers 2


Yes, for the Newtonian version of gravity (a field is sourced from mass, i.e:

$$ \vec \nabla \cdot \vec F_G \propto \rho $$

which in one dimension reads:

$$ \frac{dF}{dx} = \rho $$

For a point source:

$$ \frac{dF}{dx} = m\delta(x) $$

so $$ F(x) = \Theta(x) + C $$

Note if you use Gauss's Law to compute the (3D) field from a sheet of mass (or charge, for the electric field), you do get a constant field on each side of the sheet.


We could do some calculations: -

The "field equations" (Newton's gravitational theory is not actually a field theory, but there is no harm in considering his theory to be a classical field theory) of Newtonian gravity is given by -

$$\nabla \cdot \vec{g} = -4\pi G\rho$$

Where $\vec{g}$ is a vector field which represents the gravitational field.

For a point mass in one dimension, we get the following -

$$\rho = M\delta(x)$$

Where $\delta(x)$ is the dirac delta function and $M$ is the mass of the body. Also because we are dealing with one dimension, the gravitational field will only have an $x$ component and it can be thus described as a scalar field, which we will call $g_x$. The equations reduce to -

$$\frac{\partial{g_x}}{\partial{x}} = -4\pi GM\delta(x)$$

Which can be re-written as -

$$\int \frac{\partial{g_x}}{\partial{x}} dx = \int -4\pi GM\delta(x) dx$$

Which simplifies to - $$g_x = -4\pi GM$$

Thus, its actually true that in a one-dimensional universe, the gravitational force does not actually depend upon the distance between the bodies. Moreover, in one-dimension, the strength of gravity increases by a factor of $4\pi$. This is a very strange and pecuilar fact that arises from the field formulation of Newton's theory of gravitation.

  • $\begingroup$ The fact that strength of gravity in one-dimensional space is no longer dependent on distance, and is stronger by a factor of 4π (about 12.57 times stronger than in 3D space), it means that more mass would be clumping together, and there would be more chance of black holes forming as well. The weird thing about black holes in one dimension would be that they would pretty much divide the two sides of the "line universe", as anything attempting to cross would be eaten up by the black hole. In short, one-dimensional space with gravity appears to be quite unstable. $\endgroup$ Commented May 27 at 7:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.