# Divergence of Newton's law of gravitation

• bitrex
In summary: If you are unfamiliar with the divergence theorem, you can Google it for an explanation.In summary, the conversation discusses the calculation of the divergence of a gravitational field in cartesian and spherical coordinates. It is easier to compute in spherical coordinates, but the integral of the divergence over a volume is -4pi mG, leading some to write the divergence as a delta function. The proof for this is through the use of the divergence theorem and evaluating the surface integral.
bitrex
I am studying vector calculus, and I saw the following result in a physics text:

$$g = -\frac{m}{r^3}\vec{r}$$
$$r^2 = x^2 + y^2 + z^2$$
$$\vec{r} = ix + jy +kz$$

$$\nabla \cdot g = 0$$

I'm not sure how this was done. Is the product rule used somehow? What happened to the extra power of r? Thanks for any advice.

It helps to use the spherical version of divergence.
=

I don't expect you to know the spherical version (few people know it by heart), but doing that problem in cartesian coordinates is only for masochists.

This is actually an interesting problem, though. That field is the gravitational field of a point mass (or electric field of a point charge) at the origin. Since there is a nonzero field everywhere, you'd expect there to some source of the field, i.e., a point of nonzero divergence!

You'll find that if you plug your field into that formula for the divergence, it is zero everywhere but has a value of 0/0 at the origin. In many texts, they give an argument for why that field's divergence should actually be a dirac delta function (i.e., there is an infinitely small region of nonzero, finite divergence at the origin.)

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Okey.But why don't you use the cartesian coordinates, why the spherical coordinates??and why masochism.

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Computing $\nabla\cdot \vec g$ in spherical coordinates is easy. In spherical coordinates, the gravitational acceleration is

$$\vec g= -\frac {mG}{r^2}\hat r$$

Thus with the caveat that $r\ne 0$,

$$\nabla \cdot \vec g = \frac 1 {r^2} \frac {\partial}{\partial r}\left(r^2\left(-\frac {mG}{r^2}\right)\right) = \frac 1 {r^2} \frac {\partial }{\partial r}\left(-mG\right) = 0$$The masochism arises in computing it in cartesian coordinates. Try it.

D H said:
Computing $\nabla\cdot \vec g$ in spherical coordinates is easy. In spherical coordinates, the gravitational acceleration is

$$\vec g= -\frac {mG}{r^2}\hat r$$

Thus with the caveat that $r\ne 0$,

$$\nabla \cdot \vec g = \frac 1 {r^2} \frac {\partial}{\partial r}\left(r^2\left(-\frac {mG}{r^2}\right)\right) = \frac 1 {r^2} \frac {\partial }{\partial r}\left(-mG\right) = 0$$

The masochism arises in computing it in cartesian coordinates. Try it.

But the divergence of gravitational acceleration could be $$-4\pi.GM$$ isn't it.

coki2000 said:
But the divergence of gravitational acceleration could be $$-4\pi.GM$$ isn't it.
You are thinking of the integral of the divergence over some volume,

$$\iiint_V \nabla \cdot \vec g \, dV = \oint_S \vec g \cdot d\vec S = -4\pi mG$$

This is why some write the divergence as a (three dimensional) delta function.

D H said:
You are thinking of the integral of the divergence over some volume,

$$\int_V \nabla \cdot \vec g \, dV = \oint_S \vec g \cdot d\vec S = -4\pi mG$$

This is why some write the divergence as a (three dimensional) delta function.

Thanks.Alright what is its proof that $$\int_V \nabla \cdot \vec g \, dV = \oint_S \vec g \cdot d\vec S = -4\pi mG$$? Do you explain to me?

The first equality is the divergence theorem. The second equality results from evaluating the surface integral.

## What is the divergence of Newton's law of gravitation?

The divergence of Newton's law of gravitation refers to the discrepancy between the predictions of Newton's law and the observations of astronomical phenomena, such as the orbit of Mercury, that cannot be fully explained by the law.

## Why is there a divergence in Newton's law of gravitation?

The divergence is primarily due to the fact that Newton's law is based on the assumption of a static, homogeneous, and isotropic universe, which does not accurately reflect the complexities of the universe. Additionally, it does not take into account the effects of relativity and quantum mechanics.

## How does Einstein's theory of general relativity address the divergence?

Einstein's theory of general relativity provides a more comprehensive and accurate explanation of gravity by incorporating the concept of spacetime curvature and the effects of mass and energy on it. This theory has been able to successfully explain the discrepancies observed in the orbits of planets and other astronomical phenomena.

## What evidence supports the divergence of Newton's law of gravitation?

The most prominent evidence for the divergence of Newton's law is the precession of Mercury's orbit, which cannot be fully explained by the law and requires the use of general relativity. Other evidence includes the bending of light by massive objects, the gravitational redshift, and the observation of gravitational waves.

## Is Newton's law of gravitation still useful despite the divergence?

Yes, Newton's law of gravitation is still useful for making predictions in most situations and is still the basis for many calculations in classical mechanics. However, it is not sufficient to fully explain the behavior of gravity in extreme cases or on a cosmic scale, where the effects of general relativity become more significant.

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