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The elevator thought experiment states that one cannot distinguish between an observer inside an accelerating elevator and one inside a stationary elevator within a gravitational field.

Why does this imply something about inertial mass? At which point would the two scenarios become distinguishable if inertial mass were not equal to gravitational mass? The equivalence seems to be motivated by $F = m\cdot a$, but the experiment does not mention any force being measured.

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  • $\begingroup$ Replace the gravitational mass with the electrical charge in a uniform electric field. Then try to cancel the corresponding force by changing reference frame. $\endgroup$ Jan 23, 2022 at 20:39

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Inertial mass and gravitational mass are identical, both of them being identical to mass-energy (as opposed to the now widespread convention that mass is the invariant mass - the sum of the rest masses of all the particles in a sample). That is, inertial mass and gravitational mass are both the $m$ in $E=mc^2$.

Nothing complicated is required to measure a difference between inertial and gravitational mass. The experiment is simple enough to do it in a high school setting.

Measure an object's weight $F_g$ with a scale. Drop the object and see how fast it accelerates in gravity $g$.

$m_{g}=F_g/g$

Put the object on a cart on a horizontal track with low-friction wheels. Use an Atwood machine with a lightweight string to accelerate the cart at some fixed acceleration $a$. Measure the object's inertial weight $F_i$ with the same scale turned 90 degrees and re-zeroed in the accelerated state.

$m_i = F_i / a$.

If $m_i \ne m_g$ then you've either made an experimental error or you've just earned a Nobel prize.

The mind-straining implications of the equivalence principal have to do with things other than the equivalence of inertial and gravitational mass, which is a simple and easily verified fact.

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  • $\begingroup$ I don't doubt that the equivalence has been tested countless times, but your answer is unrelated to the thought experiment. Einstein said that the equivalence is not a coincidence. However, his thought experiment has no apparent connection to inertial mass if it does not involve a force measurement, as yours had with the Atwood machine. $\endgroup$
    – Damian
    Jan 23, 2022 at 21:21
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Suppose an accelerated rocket in outer space, with the engines regulated such that a loose object falls with acceleration $g$ in the rocket frame. When the same object is hanging from a string, the deflection indicates a force $F$. The relation $$m_i = \frac{F}{g}$$ is the inertial mass, because the object acceleration (for an inertial frame) is proportional to the force of the spring, according to Newton's second law.

Now the same object is hanging from a spring at the surface of the earth. According to the principle of equivalence, the spring deflection is the same, indicating the same force $F$. So $$m_g = \frac{F}{g} = m_i$$

We could imagine a different outcome, so it is an experimental result.

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  • $\begingroup$ Your experiment made me realize that there are two types of $m_g$'s. Because of the GR context, I only focused on the active type, the one which bends spacetime, not the "passive" one which just reacts to other masses. The $m_g$ you pointed out is of the passive type as it reacts to earths gravitation. Now I wonder if GR even has a notion of passive gravitational mass. If free fall is "following spacetime curvature", there seems to be no need for a passive $m_g$. $\endgroup$
    – Damian
    Jan 23, 2022 at 22:52
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let me also try to answer simply.

In principle inertial mass and gravitational mass are two completely different quantities. Inertial mass is the property of an object to maintain its state of motion, namely its velocity, and it's the mass that appears in Newton's law.

$$ \vec{F} = m_I \vec{a} $$

On the other hand, gravitational mass appears in the law of gravity

$$ \vec{F}(r) = G\frac{m_GM_G}{r^2}\hat{r} $$

and it's pretty much the same as charge in Coulomb's law (if we disregard the signs). In principle the two things are different exactly how, for a charged particle, (inertial) mass and charge are different.

Now, being unable to distinguish an accelerated frame from a gravitational field is equivalent to $m_I = m_G$. In fact, let's say we have a group of masses $m^i_G$ in the gravitational field produced by $M_G$ (locally the field has parallel vectors), at some distance R. These masses will "feel" the forces $\vec{F}^i = G\frac{m^i_G M_G}{R^2}\hat{r}^i$.

If $m^i_I = m^i_G$ then every mass will accelerate with acceleration

$$ \vec{a}^i = G\frac{m^i_G M_G}{m^i_I R^2}\hat{r}^i = G\frac{M_G}{R^2} \equiv \vec{g} $$

which is exactly what we expect to see in an accelerated frame. The reciprocal attractions between the masses would be the same in the two scenarios, therefore if $m_I = m_G$ the accelerated frame and gravitational field are perfectly equivalent and there is no measurement we can perform in order to decide in which scenario we are.

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  • $\begingroup$ "If $m^i_I = m^i_G$ then every mass will accelerate with acceleration $\vec{a}^i$": Shouldn't they accelerate with the same rate regardless of $m^i_I = m^i_G$? $\endgroup$
    – Damian
    Jan 23, 2022 at 22:12
  • $\begingroup$ well, each mass would be subject to a force $\vec{F}^i \propto m^i_G$ and would accelerate with acceleration $\vec{a}^i \propto \frac{m^i_G}{m^i_I}$ where $\propto$ means "proportional". As you can see, the accelerations will depend on the ratio $\frac{m^i_G}{m^i_I}$ $\endgroup$
    – Andrea
    Jan 24, 2022 at 11:15

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