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!)


3 Answers 3


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.


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.


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.


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