The Newtonian Gravitational Field and Potential

A Newtonian potential is not itself completely observable. The acceleration field $\mathbf g=-\nabla\Phi$, potential differences, and tidal derivatives are invariant under $\Phi\mapsto\Phi+C$; moreover, a uniformly accelerating frame can absorb a term linear in position. Once a frame and boundary normalization are fixed, $\Phi$ efficiently encodes gravitational work and motion.

Required background. Calculus and Linear Algebra provides multivariable differentiation, linear maps, inner products, and basis changes. Helpful background. Ordinary Differential Equations provides initial-value problems, linearization, and asymptotic solution methods.

Field, potential, and free fall

For a test body with inertial mass $m_{\rm i}$ and gravitational mass $m_{\rm g}$,

$$m_{\rm i}\ddot{\mathbf x}=-m_{\rm g}\nabla\Phi.$$

Universality of free fall is the empirical statement that $m_{\rm g}/m_{\rm i}$ is body-independent. Choosing mass units so that the ratio is one gives $\ddot{\mathbf x}=\mathbf g$ with

$$\mathbf g=-\nabla\Phi.$$

For a stationary potential, the work per unit gravitational mass from $A$ to $B$ is

$$\frac{W_{A\to B}}{m_{\rm g}} =\int_A^B\mathbf g\cdot d\boldsymbol\ell =-\bigl[\Phi(B)-\Phi(A)\bigr].$$

Thus potential differences, not the zero of potential, determine conservative work. For a time-independent field, $E=m v^2/2+m\Phi$ is conserved. These relations, the point-mass potential, and the universality of gravitational free fall are developed in Tong 2013, §§2.2–2.3.3, pp. 18–26, PDF.

If $\Phi(t,\mathbf x)$ changes during the journey, the chain rule instead gives

$$\frac{W}{m} =-\bigl[\Phi(t_B,\mathbf x_B)-\Phi(t_A,\mathbf x_A)\bigr] +\int_{t_A}^{t_B}\partial_t\Phi(t,\mathbf x(t))\,dt.$$

Here inertial and gravitational masses have been identified. Equivalently, $d(v^2/2+\Phi)/dt=\partial_t\Phi$ along a freely falling trajectory. An evolving source can exchange energy with the test body; a spatially conservative force at each instant need not conserve mechanical energy over time. A purely conventional $C(t)$ shifts both potential terms in this identity and leaves the actual work unchanged.

What changes with convention or frame

Adding a spatial constant $C(t)$ leaves $\mathbf g$ unchanged. A term linear in position has two distinct interpretations. Within one fixed spatial chart, the active replacement

$$\widetilde\Phi(t,\mathbf x) =\Phi(t,\mathbf x)+\mathbf A(t)\cdot\mathbf x+C(t)$$

changes the field by the spatially uniform vector $-\mathbf A(t)$ and therefore changes trajectories in that fixed chart. By contrast, consider the passive change to translating coordinates $\mathbf x'=\mathbf x-\mathbf X(t)$, so that $\mathbf x=\mathbf x'+\mathbf X(t)$. Because

$$\ddot{\mathbf x}' =\mathbf g\!\left(t,\mathbf x'+\mathbf X(t)\right)-\ddot{\mathbf X}(t),$$

the potential measured in the accelerating coordinates is

$$\Phi'(t,\mathbf x') =\Phi\!\left(t,\mathbf x'+\mathbf X(t)\right) +\ddot{\mathbf X}(t)\cdot\mathbf x'+C(t).$$

It generates

$$\mathbf g'(t,\mathbf x') =\mathbf g\!\left(t,\mathbf x'+\mathbf X(t)\right)-\ddot{\mathbf X}(t).$$

The translated argument is essential: the active fixed-chart shift and the passive coordinate transformation are not the same operation. At corresponding physical points, $\partial'_i\partial'_j\Phi'=\partial_i\partial_j\Phi$; a uniform frame acceleration can remove a uniform field locally but cannot remove tidal effects.

For an isolated mass distribution, the boundary condition $\Phi\to0$ as $r\to\infty$ fixes the additive constant. A point mass $M$ then has

$$\Phi(r)=-\frac{GM}{r}, \qquad \mathbf g=-\frac{GM}{r^3}\mathbf r.$$

The sign follows from attraction: $d\Phi/dr=GM/r^2\gt0$, so $-\nabla\Phi$ points inward. Dimensions give $[\Phi]=L^2T^{-2}$ and $[G]=L^3M^{-1}T^{-2}$.

For a continuous density,

$$\Phi(\mathbf x)=-G\int \frac{\rho(\mathbf y)}{|\mathbf x-\mathbf y|}\,d^3y,$$

provided the source is sufficiently localized and the potential is normalized at infinity. Superposition is exact because Newtonian field theory is linear.

This integral also exposes the test-body approximation. The source $\rho$ is treated as prescribed; the test body’s own field and its backreaction on the source are neglected. For two comparable bodies, one must solve their coupled motion even though the same pair potential applies. Extended bodies introduce another qualification: the center of mass follows the point-particle field only when the external field varies negligibly across the body or when internal multipoles can be ignored. The potential is therefore a field on space, not merely “potential energy divided by mass” for every composite object without approximation.

A shell as a discriminating example

For a thin spherical shell of mass $M$ and radius $R$, spherical symmetry and Gauss’s law give

$$\mathbf g=\mathbf0 \quad (r\lt R), \qquad \mathbf g=-\frac{GM}{r^2}\mathbf e_r \quad (r\gt R).$$

Continuity of $\Phi$ at the shell and $\Phi(\infty)=0$ yield

$$\Phi(r)= \begin{cases} -GM/R, & r\lt R,\\ -GM/r, & r\gt R. \end{cases}$$

The interior potential is nonzero yet produces no force. This separates “potential value” from observable acceleration. A particle crossing the shell also has continuous potential energy, while the field jumps because the idealized surface density is singular.

As a second check, take a circular orbit in the point-mass field. The inward acceleration condition $v^2/r=GM/r^2$ gives $v^2=GM/r$. Its specific energy is $v^2/2-GM/r=-GM/(2r)\lt0$, confirming that the sign convention makes a circular Kepler orbit bound. Shifting $\Phi$ by a constant shifts this numerical energy but changes neither the orbit nor the energy difference required to escape.

Common pitfalls

Confusing gauge freedoms. The additive constant is always irrelevant to forces. A linear term is removable only together with a change to an accelerating frame; within one fixed inertial frame it changes trajectories.

Treating Newtonian gravity as causal. Poisson theory determines the field instantaneously from the density on the same absolute-time slice. It is a weak-field, slow-motion limit, not a propagation theory.

Exercise: escape from a point mass

Find the minimum launch speed at radius $R$ needed to reach infinity with zero final speed.

Solution

With $\Phi(\infty)=0$, conservation of specific energy gives

$$\frac12v_{\rm esc}^2-\frac{GM}{R}=0,$$

and hence $v_{\rm esc}=\sqrt{2GM/R}$. The argument requires a stationary point-mass field and ignores drag, other bodies, and relativistic corrections. The result is dimensionally a speed and increases with $M/R$.

Poisson’s Equation and Gravitational Boundary Data makes precise how density and boundary conditions jointly determine $\Phi$.

References

• Tong, David. Dynamics and Relativity. University of Cambridge lecture notes, 2013. §§2.2–2.3, pp. 18–26, PDF.