4
$\begingroup$

We are familiar with Newton's law of gravitation:

$$\textbf{F} = \frac{-GMm}{r^2} \hat{\textbf{r}},\tag{1}$$

which leads to a gravitational field strength relation:

$$\textbf{g} = \frac{-GM}{r^2} \hat{\textbf{r}}. \tag{2}$$

In terms of vector calculus we can write this in the form:

$$\nabla\cdot\textbf{g} = -4\pi G \rho \tag{3}$$

(where $\rho$ is the mass density) in analogy with Coulombs law and Maxwells first law.

My question is whether this equation is sufficient to fully describe (Newtonian) gravitation, or whether a relation for the curl, $$\nabla \times \textbf{g} = 0,\tag{4} $$ is also required?

If so, it seems unusual to me that Newton's force law requires just one equation, yet a vector calculus approach would require two. (But maybe that's just the way it is!)

$\endgroup$

3 Answers 3

4
$\begingroup$

Yes, the Newtonian gravitational field ${\bf g}$ is also required to be rotation-free $\nabla \times {\bf g} = 0$. This also follows from the existence of a Newtonian gravitational potential.

$\endgroup$
2
$\begingroup$

You can show that if there exists a scalar potential for the field, it will also be curl-free. So the relation is required, but stating it separately is redundant, as it is a consequence of the existence of the potential.

$\endgroup$
0
$\begingroup$

Yes, you also need

$$\nabla \times \mathbf{g} = \mathbf{0}$$

This is exactly analogous to the situation in electrostatics, where Gauss's law

$$\nabla \cdot \mathbf{E} = \frac{\rho_q}{\epsilon_0}$$

is insufficient to give the electric field, and we also need

$$\nabla \times \mathbf{E} = \mathbf{0}$$

(together with some assumptions about the underlying "vacuum" [i.e. no radiation is present])

which itself is a special case of Faraday's law

$$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$$

when the magnetic field remains unchanged with time. In the case of Newtonian gravitation, there is no "gravimagnetic field", as all changes propagate instantaneously, so the right-hand side of the gravitational equation is always zero.

$\endgroup$

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.