# Why does proper acceleration equal the coordinate acceleration measured in an instantaneous inertial rest frame?

Consider an arbitrarily moving particle in special relativity. In a general inertial frame of reference, we can measure the coordinate velocity $$v$$ and acceleration $$a$$ of the particle. In an instantaneous inertial rest frame, i.e. in a coordinate frame as determined by a non-accelerating observer with equal velocity $$v$$ as the particle at an instant, the particle has coordinate velocity $$v=0$$ and coordinate acceleration $$a$$. Seemingly all sources I have consulted now state that this coordinate acceleration, which happens to equal the spacial part of the acceleration four-vector of the particle, corresponds to the acceleration the particle 'feels', i.e. the acceleration that an observer accelerating with the particle would measure on an accelerometer carried with the particle, which they define to be the particles proper acceleration. Why is this the case? Why does proper acceleration of the particle equal the coordinate acceleration of the particle in an instantaneous rest frame? Is this to be understood as an experimentally verified postulate akin to the clock hypothesis, which states that the time measured on an ideal clock carried by the accelerated particle is equal to the integral $$\tau = \int \sqrt{1 - v(t)²/c²} \mathcal{d}t$$ defining proper time?

In the particle's original rest frame, once the particle has undergone any degree of acceleration at all, relativistic effects would come into play and they would get more and more significant as the particle's speed increased, so that the apparent acceleration in that original frame would bear no resemblance to the acceleration experienced by the particle. It is only in the particle's instantaneous rest frame that relativistic effects are entirely negligible.

To see this, imagine you are in car travelling at almost c with respect to some observer. In your frame you can press the gas pedal and accelerate as normal, whereas to the observer you would experience virtually zero acceleration.