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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.
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.
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.
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.
