# Interpretation of finite proper time of fast-accelerating objects for infinite coordinate time

Say we have an object with velocity in earth's frame that accelerates quickly.

For example, suppose we have some object that accelerates very quickly, and has speed $$v$$ in earth's frame of the form

$$v(t)=c\sqrt{1-e^{-2t}}$$.

Then the proper time from earth's time when it starts accelerating, $$t=0$$ to some other earth time $$t=b$$ is

$$\tau (b)=\int_0^b\sqrt{1-\frac{v(t)^2}{c^2}}dt=\int_0^b\sqrt{1-\sqrt{1-e^{-2t}}^2}dt=\int_0^be^{-t}dt$$

$$\tau (b) = 1-e^{-b}$$.

And so $$\lim_{b\rightarrow \infty}\tau (b)=1$$.

In another example, suppose $$a$$ is some constant, and

$$v(t)=c\text{Tanh}(\frac{at}{c})$$.

Then $$\lim_{b\rightarrow\infty}\tau(b)=\lim_{b\rightarrow\infty}\int_0^b\sqrt{1-\frac{v(t)^2}{c^2}}dt=\lim_{b\rightarrow\infty}\int_0^b\sqrt{1-\text{Tanh}(\frac{at}{c})^2}dt=\lim_{b\rightarrow\infty}\int_0^b\text{sech}(\frac{at}{c})dt$$

$$\lim_{b\rightarrow\infty}\tau(b)=\frac{c\pi}{a}$$.

So there are a few cases here of a physical object that never reaches $$c$$ in earth's frame, but has a finite proper time for infinite coordinate time.

I am wondering what exactly is happening in the frame of the object right as this proper time passes, and after this proper time passes?

What you've discovered is that in the rest frame of the accelerating observer there is a coordinate singularity. Indeed this is closely related to the event horizon in the Schwarzschild geometry, and understanding one can help you understand the other.

If we do a transformation into the rest frame of the accelerating observer the spacetime geometry in that frame is described by the Rindler metric:

$$ds^2 = -\left(1 + \frac{ax}{c^2}\right)^2 c^2 dt^2 + dx^2 + dy^2 + dz^2 \tag{1}$$

where $$a$$ is the proper acceleration of the accelerating observer (note that in this frame $$a$$ is negative). We'll take the observer to be accelerating in the $$x$$ direction so $$dy=dz=0$$. For a light beam $$ds=0$$ so in this geometry the equation of motion for a light beam moving in the $$x$$ direction is obtained from (1) by setting $$ds=0$$:

$$\frac{dx}{dt} = \left(1 + \frac{ax}{c^2}\right) c \tag{2}$$

Compare this with the equation we get if we use the same method on the Schwarzschild metric:

$$\frac{dr}{dt} = \left(1 - \frac{2GM}{c^2r}\right)^{1/2} c \tag{3}$$

And there is an obvious similarity. Indeed, suppose we take equation (3) and consider large $$r$$ where the gravitational acceleration is approximately given by Newton's law:

$$a = -\frac{GM}{r^2}$$

Substituting this into (3) gives:

$$\frac{dr}{dt} \approx \left(1 + \frac{2ar}{c^2}\right)^{1/2} c$$

and then using a binomial expansion to approximate the square root gives:

$$\frac{dr}{dt} \approx \left(1 + \frac{ar}{c^2}\right) c$$

And remarkably we find the accelerating observer in flat space and the observer near a black hole get the same equation of motion for light. Specifically both observers find that the velocity of light falls to zero at a certain point. For the Schwarzschild observer this is of course the event horizon at $$r_s = 2GM/c^2$$ and for the accelerating observer it is the Rindler horizon at $$x = c^2/a$$.

Notoriously, objects falling into a black hole take infinite coordinate time but finite proper time to reach the horizon, and we get the same result for the accelerating observer. If the accelerating observer drops an object then that object also takes infinite coordinate but finite proper time to reach the horizon.

But the dropped object is of course just an object stationary in flat spacetime, watching the accelerating observer speed away, so there is no horizon present (just as for an observer falling freely into a black hole no horizon is present). In both cases the singularity is just a place where the transformation between the two coordinate systems becomes singular.

Note, that for such a trajectory the acceleration in its instantaneous rest frame would be diverging (as $$b\to \infty$$), which means that no physical object could move along such trajectory indefinitely, it would be destroyed by infinite forces applied to it. Such trajectory is not extendable past this point.

If we try to interpret this situation in the context of general relativity, and assume that an object of finite mass moves along such a trajectory then the momentum that one needs to transfer to this object would be diverging in all reference frames, and if we take the backreaction into account, any force that could produce such a trajectory would also produce a true curvature singularity.

One could contrast such behavior with a milder case of constant proper acceleration. Though the total energy of an object undergoing such an acceleration would be diverging with time in the frame of any inertial observer, in the frame of an accelerating object (Rindler frame) the object would experience only finite stresses and so it could exist there for all values of its proper time, which would be growing logarithmically relative to the time of inertial observers. If we include the backreaction into account we could obtain GR solutions such as C-metric (see this paper for an overview), representing a finite mass (a black hole) undergoing constant proper acceleration.