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In classical physics velocity and acceleration follow: $v = at$, so an object falling at a constant $10 m/s^2$ for 50 seconds would wind up having a velocity of $500 m/s$.

However, according to this the same object falling for 312 days would reach 90% of the speed of light, and pass it after 347 days - which is clearly wrong.

I have tried to look around, but the closest I found to something I understood was this answer which states that the energy required to accelerate something follows this equation:

$$ p = \text{“percent of speed of light”} $$ $$ E = \gamma mc^2 = \frac{mc^2}{\sqrt{1 - p^2}} $$

So, at this point a naive approach is to say that accelerating something adds energy, and therefore maybe you can grab the $\gamma$ part and smush it onto $v = at$:

$$ v = a t \frac{1}{\sqrt{1 - p^2}} $$

With the result that it would now take:

$$ \frac{0.9c}{10\frac{m}{s^2}} \frac{1}{\sqrt{1 - 0.9^2}} \approx 716 days $$

Of constant falling/accelerating to reach 90% of the speed of light, and reaching c is impossible.

On the one hand, this sounds somewhat logical - but on the other hand, it also sounds too easy. So, am I right, close, or way off in la-la-land? Also, what is the correct answer?

Ps. Yes, this question is inspired by portal shenanigans like in the game Portal.

Qmechanic
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Frodyne
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1 Answers1

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In relativity there are two distinct concepts of acceleration. One is “coordinate acceleration”. This has the usual relationship $\vec a= \frac{d}{dt}\vec v$. This is a coordinate-dependent concept, so in that sense it is non-physical.

The other concept of acceleration is “proper acceleration”. This is the acceleration measured by an accelerometer. It is invariant, agreed-on by all observers, and directly measurable. Thus proper acceleration is the physical acceleration.

An object experiencing constant proper acceleration does not follow a parabolic path as it would in Newtonian “spacetime”. Instead it follows a hyperbolic path in relativistic spacetime. The asymptote of that hyperbola is $c$, so a constant proper acceleration will approach $c$ but never reach or exceed it.

Dale
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