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Objects with mass that continuously accelerate will eventually approach $c$, but cannot exceed it. So if I find myself in an elevator, unable to determine if I'm in a uniform gravitational field or accelerating, I will resolve this by:

  • calculating the acceleration, a
  • solve the equation for t, the amount of time it takes for my mass to reach c accelerating at a.
  • If the acceleration stops before t, I was accelerating through empty space
  • If the acceleration continues beyond t, I must be in a uniform gravitational field

Why won't this work?

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    $\begingroup$ At an acceleration that you would measure as being constant, you STILL will not reach the speed of light, even if you accelerate for your whole lifetime. The special relativity equations that govern this phenomenon are not linear (as you expect). In addition, velocity is measured RELATIVE to some other object, so velocity is not an absolute thing for your particular reference frame. If you investigate the equations that govern special relativity, and calculate velocities, time differences, etc., from those equations, you will see what I am referring to. $\endgroup$ – David White Feb 1 at 0:12
  • $\begingroup$ If I increase my velocity by 1m/s/s, how could I not approach 299,792,458 ? I understand that it takes more and more energy to actually reach that number to the point it becomes an impossibility, but why can't I use that limit to detect what's happening outside my elevator? Also, regarding RELATIVE velocity, ok suppose I was moving "backward" at (c - 1m/s), then my acceleration can go "forward" no more than 2*c. I can't be moving in any direction faster than c, so the most I can go in the "opposite" direction is just under 2.0*c $\endgroup$ – mike Feb 1 at 0:37
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    $\begingroup$ You can have constant proper acceleration forever en.wikipedia.org/wiki/Proper_acceleration, an inertial observer watching you from outside will see you have a slowing coordinate acceleration though, the slower the closer you are to c. $\endgroup$ – Wolphram jonny Feb 1 at 1:04
  • $\begingroup$ @mike, you would have to have an outside observer who never accelerated tell you your velocity. As Wolphram jonny pointed out, that observer would witness your mass get larger and larger as you gained more speed, and that outside observer would see your acceleration slow WAY down, even though you measured a constant acceleration. I say again, that you would do well to read up on special relativity, and look at the equations that are related to your question. $\endgroup$ – David White Feb 1 at 1:46
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    $\begingroup$ Editing the last sentence: Only an external and relative rest observer, can notice that the acceleration of the elevator decreases progressively. $\endgroup$ – João Bosco Feb 2 at 2:24
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In your idea, the at rest observer sees the accelerating rocket (or elevator) acquire a velocity $\Delta v_1=a \Delta t_1$ then an additional $\Delta v_2=a \Delta t_2$ in the next interval of time $\Delta t_2$. You assumed all the $\Delta v_i$ will eventually add up to c or greater. Unfortunately, according to special relativity, velocities do not add this way and instead a special velocity addition formula must be used that prevents ever reaching c.

Whereas velocities are not additive, there is a thing called the Lorentz Boost parameter (also called rapidity) $\lambda$ which is additive for the observer. It is related to the velocity by $\frac{v}{c}=tanh(\lambda)$. Notice that for an infinitesimal velocity $\frac{\Delta v}{c}= \Delta \lambda$ and then $\frac{a\Delta t}{c}=\frac{\Delta v}{c}= \Delta \lambda$ where $a$ is the constant acceleration experienced in the rocket frame and $\Delta t$ is the time that passes in the rocket frame. Now add up all these $\Delta \lambda$ to get $\lambda=\frac{at}{c}$. The final velocity is then $\frac{v}{c}=tanh(\frac{at}{c})$. Notice that as $t \rightarrow \infty$, $\frac{v}{c} \rightarrow 1$, and you don't get a velocity greater than c,

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I find myself in an elevator, unable to determine if I'm in a uniform gravitational field or accelerating,

This part is exceptionally important here. It means that the measurements are all done from the perspective of the person on the elevator. Thus the acceleration in question is the proper acceleration of the elevator passenger and the time in question is the proper time of the passenger.

calculating the acceleration, a

The passenger’s proper acceleration can be directly measured with an onboard accelerometer. Similarly the passenger’s proper time can be directly measured with an onboard clock.

solve the equation for t, the amount of time it takes for my mass to reach c accelerating at a. If the acceleration stops before t, I was accelerating through empty space If the acceleration continues beyond t, I must be in a uniform gravitational field Why won't this work?

You can do it, but the t is infinite. The passenger can go for an infinite amount of proper time at a constant proper acceleration without exceeding c. From the perspective of an external inertial observer the coordinate acceleration will decrease and time dilation will increase as the elevator asymptotically approaches c but never exceeds it.

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  1. An infinite amount of time would be required to reach c. 2.You will have to wait till infinity to actually see if the elevator stops. 3.Your reasoning would always lead you to the conclusion that you are in a uniform gravitational field. 4.If you are using a fictional device that provides this acceleration. The device has no reason to stop before 't'. 5.It won't even stop after 't'. Because its attached to your frame and there is no increment in your mass for the fictional device.
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In Newtonian kinematics a ship accelerating at one $g$ will reach the speed of lght in about one year (This the "god whale equation" $gy=c$).

However, in relativistic kinematics, the spacetime trajectory of a particle uniformly acclerating at $a$ is $$ ct= \frac{c^2}{a}\sinh (a \tau/c), \quad x= \frac {c^2} {a} \cosh (a\tau/c), $$ where $\tau$ is the elapsed proper time for someone riding with the particle. This is uniform acceleration as the hyperbolic trajectory in $(t,x)$ space in the Minkowski space analogue of a circle. Consequently an observer riding on the particle (elevator) feels the same acceleration (gravity) for all of her $\tau$. The speed observed in the $(t,x)$ coordinate system is $$ \frac{dx}{dt}= c\sinh(a\tau/c)/\cosh (a\tau/c)= c\tanh (a \tau/c) $$ so it takes infinitely long time $\tau$ for the particle (elevator) to reach the speed of light.

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This is an experiment you can actually do, provided you have an ultra high energy cosmic ray (UHECR) at the GZK limit ($E=5\times 10^{19}$ eV). For a proton ($M = 938\times 10^6$ eV), that travels at:

$$ \frac v c = \sqrt{1-\frac 1 {\gamma^2}} \approx 1-\frac 1 {2\gamma^2}=1-\frac {M^2}{2E^2}$$

or about 2.3 inches per slower than $c$.

In the reference frame of that UHECR, you are moving at almost $c$ upward.

When you push the "up" button, you still go up.

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