Inertial Mass, Gravitational Mass, and Universality

“Inertial mass equals gravitational mass” compresses three logically different quantities into one phrase. Inertial mass measures resistance to acceleration, passive gravitational mass measures response to an externally specified gravitational field, and active gravitational mass measures how strongly a body sources that field. Universal free fall directly tests only a ratio involving the first two. Keeping the three roles separate prevents an empirical statement about test-body motion from being mistaken for a complete theory of gravitational sources.

Required background. We use Minkowski geometry and relativistic normalization, the distinction between coordinate and observer-measured components, and the Newtonian gravitational potential.

Helpful background. Tensor fields and natural operations clarify the later geometric reformulation, while proper acceleration and accelerometer readings help distinguish free fall from supported motion.

Three roles played by mass

For a slowly moving test body in a prescribed Newtonian potential $\Phi$, write

$$m_{\mathrm I}\mathbf a=-m_{\mathrm P}\boldsymbol\nabla\Phi.$$

The coefficient $m_{\mathrm I}$ is inertial mass and $m_{\mathrm P}$ is passive gravitational mass. The field itself is sourced by active gravitational mass. For separated point sources one may encode that role in

$$\nabla^2\Phi=4\pi G\rho_{\mathrm A}, \qquad \Phi=-\frac{Gm_{\mathrm A}}{r}$$

outside a spherical source of active mass $m_{\mathrm A}$. These definitions concern different experiments: push the body to determine $m_{\mathrm I}$, place it in a calibrated external field to infer $m_{\mathrm P}$, and infer $m_{\mathrm A}$ from the field it produces.

There is a conventional normalization hidden here. Rescaling every passive mass by a common factor and inversely rescaling $\Phi$ leaves test-body motion unchanged; rescaling active mass can similarly be absorbed into $G$. Equality of numerical values therefore becomes meaningful only after one common normalization of force, field, and $G$ has been fixed. Reall’s Newtonian discussion makes exactly this separation before passing to relativistic universality Reall 2013, § 1.2, pp. 2–3, PDF.

What universal free fall measures

Dividing the equation of motion by $m_{\mathrm I}$ gives

$$\mathbf a=-\left(\frac{m_{\mathrm P}}{m_{\mathrm I}}\right)\boldsymbol\nabla\Phi.$$

Suppose bodies $A$ and $B$ are released at the same event with the same velocity, are small enough not to disturb the external field, and differ only in composition. Along a chosen sensitive direction $\hat{\mathbf n}$,

$$\Delta a_{AB} =-\left[ \left(\frac{m_{\mathrm P}}{m_{\mathrm I}}\right)_A -\left(\frac{m_{\mathrm P}}{m_{\mathrm I}}\right)_B \right]\hat{\mathbf n}\!\cdot\!\boldsymbol\nabla\Phi.$$

Thus a vanishing differential acceleration constrains equality of passive-to-inertial ratios. Once the common ratio is absorbed into the definition of $\Phi$, one conventionally writes $m_{\mathrm P}=m_{\mathrm I}$. The observation does not separately determine either mass, and it says nothing yet about $m_{\mathrm A}$.

A dimensionless differential-acceleration parameter is often defined by $\eta_{AB}=2(a_A-a_B)/(a_A+a_B)$ after projecting signed accelerations along the same direction. Its value is meaningful only when the two bodies sample the same external field and have matched initial data. Will’s operational review carefully identifies this composition dependence as the test domain rather than quoting equality as an axiom Will 2014, § 2.1.1, pp. 7–9, PDF.

Active mass requires another argument

The source role enters a two-body force law. In a simplified isolated system, the force of body $A$ on body $B$ is proportional to $m_{\mathrm A}^{A}m_{\mathrm P}^{B}$, whereas the opposite force contains $m_{\mathrm A}^{B}m_{\mathrm P}^{A}$. If ordinary momentum conservation and equal-and-opposite internal forces are also assumed, then

$$m_{\mathrm A}^{A}m_{\mathrm P}^{B} =m_{\mathrm A}^{B}m_{\mathrm P}^{A}.$$

This implies that $m_{\mathrm A}/m_{\mathrm P}$ is universal across the bodies in that model. It is an additional inference using source dynamics and momentum conservation, not the result of a drop experiment alone. After fixing normalization, all three masses may then be represented by one parameter.

The assumptions in that inference are essential. The unequal-ratio two-body equations can instead be rewritten with effective inertial coefficients so that they possess a variational formulation and a conserved, reweighted momentum. Galilean symmetry and mathematical consistency therefore do not by themselves select the usual inertial-mass momentum or prove $m_{\mathrm A}=m_{\mathrm P}$. Giulini develops this counter-analysis explicitly Giulini 2026, §§ I–III, pp. 1–4, PDF. Equality of active and passive mass remains an empirical statement once the operational definitions of force and momentum are fixed.

General relativity ultimately replaces these Newtonian source labels by stress-energy and geometry. Even there, the lesson survives: motion of a negligible probe and generation of a gravitational field are distinct questions. Strongly self-gravitating bodies probe gravitational binding energy and lie outside the elementary test-body statement.

Regime and independent checks

The derivation assumes a weak, slowly varying external field, negligible self-gravity, negligible drag and electromagnetic forces, and bodies small compared with the field-variation scale. A charged particle supplies a useful countercheck: in an electric field its acceleration depends on $q/m_{\mathrm I}$, so different compositions need not follow the same trajectory. The cancellation of mass is special to a universal coupling, not a generic feature of forces.

Dimensional analysis provides another check. Since $\Phi$ has dimensions of squared velocity, $\boldsymbol\nabla\Phi$ has dimensions of acceleration; $m_{\mathrm P}/m_{\mathrm I}$ must be dimensionless. Finally, setting the two ratios equal makes $\Delta a_{AB}$ vanish without imposing any relation involving active mass, confirming the logical boundary of the experiment.

Exercises

Two test bodies have ratios $(m_{\mathrm P}/m_{\mathrm I})_A=1+\epsilon_A$ and $(m_{\mathrm P}/m_{\mathrm I})_B=1+\epsilon_B$, with $|\epsilon_A|,|\epsilon_B|\ll1$. Find $\eta_{AB}$ to first order when both fall along the same uniform field.

Solution

The signed accelerations have a common factor $a_0$ and are $a_A=a_0(1+\epsilon_A)$ and $a_B=a_0(1+\epsilon_B)$. Hence

$$\eta_{AB}=\frac{2(\epsilon_A-\epsilon_B)}{2+\epsilon_A+\epsilon_B} =\epsilon_A-\epsilon_B+O(\epsilon^2).$$

Only the difference is observable in this comparison; a common shift can be absorbed into the field normalization.

References

• Giulini, Domenico. “What If Active and Passive Gravitational Masses Were Not Equal?” arXiv:2607.02614v1 (2026). arXiv.
• Reall, Harvey S. General Relativity. Cambridge Part III lecture notes (2013). PDF.
• Will, Clifford M. “The Confrontation between General Relativity and Experiment.” Living Reviews in Relativity 17, 4 (2014). DOI; arXiv.