Can't ' Gravitational mass equals inertial mass ' be proved as below? :

A particle is falling in a uniform gravitational field $g$ experiencing a force $m_gg$ ($m_g$ is its gravitational mass)

In its own frame, it is not accelerating and hence the sum of all forces should be zero.

But to apply Newton's law in this frame we have to add a pseudo force $-m_ig$ where ($m_i$ is the inertial mass), and we have


$$\Rightarrow m_i=m_g$$

  • 2
    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/267741 $\endgroup$ Feb 18, 2021 at 17:01
  • 1
    $\begingroup$ Try the same argument but with the force coming from an electric field instead of a gravitational one. $\endgroup$ Feb 18, 2021 at 17:21
  • $\begingroup$ Also read one of Einstein's original papers on general relativity where the claim is made explicit by similar reasoning. Nevertheless the proof is NOT theoretical but experimental (up to current precision). Theoreticaly it can be shown that intertial and gravitational mass have a constant ratio (which by suitable units can be made 1) $\endgroup$
    – Nikos M.
    Feb 18, 2021 at 18:28
  • $\begingroup$ Why did you assume that the acceleration of the object was g? That can only be true of the gravitational and inertial masses are equal, so, as pointed out by Andrew, you assumed what you want to prove in order to prove it $\endgroup$ Feb 18, 2021 at 20:54
  • $\begingroup$ @Andrew Steane, I get an obvious equation, the left and right hand sides are the same. $\endgroup$
    – Kashmiri
    Feb 19, 2021 at 4:46

1 Answer 1


I wouldn't call your argument a "proof" in the sense that you've assumed what you want to prove.

What could have happened is that the gravitational acceleration, $g$, was not universal but depended on the gravitational and inertial mass. To give a very simple model, suppose that \begin{equation} F = m_g g(m_g,m_i) = m_i g_{E} \left(\frac{m_g}{m_i}\right)^2 \end{equation} where $g_E=9.8\ {\rm m/s}$ is the normal gravitational acceleration of Earth. Note in my notation, $g(m_g,m_i)$ is a function of the inertial mass $m_i$ and gravitational mass $m_g$ (with units of acceleration), while $g_E$ is just a number. On the right hand side, $g_E$ multiplies the square of the ratio of the gravitational and inertial masses (this is not true in Nature, it is just a silly model to prove a point).

Then using your argument, we would find \begin{equation} F = m_i a \implies a = g_E \left(\frac{m_g}{m_i}\right)^2 \end{equation} In our hypothetical reality we could imagine that, say, feathers have $m_g/m_i=1$ and lead has $m_g/m_i=2$, then a pound of lead would fall four times faster than a pound of feathers (more precisely, an intertial-pound of lead would accelerate under the influence of gravity 4 times faster than an inertial-pound of features).

The non-trivial thing in gravity is that $g(m_g,m_i)=g_E$ exactly (as far as we can tell), for all objects. Fundamentally this is an experimental fact, and not something you can prove mathematically, but it has deep theoretical implications.

Your argument is not a proof because you assumed that $g(m_g,m_i)=g_E$ from the beginning, but this is precisely the thing you would need to prove, if you could prove inertial and gravitational mass are the same (but again, you can't prove this).


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