# Acceleration of objects with speeds comparable to $c$. Can we find acceleration of the frame by treating $v$ in $1/\sqrt{1-v^2/c^2}$ as variable?

We know that the $v-t$ graph for constant acceleration is a straight line. But when the speed becomes comparable to $c$, the speed of light, relativistic effects come into play. Suppose if a rocket on earth starts from rest (A) with constant acceleration or (B) with increasing acceleration. Since, there is acceleration in both the cases speed will increase but it will never reach $c$ which means that acceleration has to decrease from somewhere. Then the graphs looks like these: I tried to obtain an expression for the acceleration by using the inverse Lorentz transformation equations in STR.
In STR the $v$ in $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is constant. But in this rocket example this $v$ is itself changing and so I treated it as a variable. Let the rocket be $S'$ and the earth be $S$ and $S'$ be moving along $x$-direction. Now to see from $S$ how $S'$ is accelerating, I took a fixed position in $x'$-axis. For simplicity, I took its origin i.e., $x'=0$ so that $x=vt'\gamma$. Then I differentiated $x$ with respect to $t$, substituted $t'$, substituted $\frac{dt'}{dt}$ and tried to get an expression for $\frac{dv}{dt}$. But the result I got does not make sense.
I am not familiar with GR so I thought this must be explained in GR and I searched in the Internet if there is any such formula but did not find any. Is it the wrong way I am going? If it is there in GR to which I am not familiar, I seek a brief explanation of how acceleration is understood in GR, if possible.

## 1 Answer

First a comment that you don't need GR for problems like this: SR is fine for any kind of accelerated motion in flat spacetime.

Imagine the motion as a sequence of boosts. Think of it at first discretely. At each step, the object undergoes the same boost: let's make it a small one of relative velocity $\Delta v$, with a boost matrix $\Lambda(\Delta v)$. So after $n$ of these boosts, we can see that the overall boost is of the form $\Lambda(\Delta v)^n$; we let the $\Delta v\to 0$ and $n$ become very large, and thus our overall boost matrix after $n$ steps looks like $\exp(n \Delta\eta)$, where you "calibrate" $\Delta\eta$ for the right acceleration. Written out in full our overall boost is:

$$\exp(n \Delta\eta) = \left(\begin{array}{cc}\cosh(n \Delta\eta)&\sinh(n \Delta\eta)\\\sinh(n \Delta\eta)& \cosh(n \Delta\eta)\end{array}\right)$$

Now suppose the accelerated particle makes the same $\Delta v$ boost after every $\Delta t$ seconds of its time. Each $\Delta t$ of its time is $\Delta t \gamma=\Delta t \cosh(n \Delta\eta)$ of "our" time: we being the observers watching the uniformly accelerating object and plotting its speed. The total time by our clock after $n$ steps is therefore proportional to $\sinh(n\,\Delta \eta) = \int_0^{n\,\Delta\eta}\cosh(u)\,\mathrm{d}u$. Thus:

$$t = k\,\sinh(n\,\Delta \eta)=\frac{k\,v}{\sqrt{1-\frac{v^2}{c^2}}}$$

(using the relationship $v/c = \tanh(n\,\Delta\eta)$) and on rearranging we get:

$$v=\frac{c\, t}{\sqrt{c^2 k^2+t^2}}$$

which for small times is $v = t/k$, so see that $k$ is the reciprocal of the acceleration. The curve you plot is therefore like your curve (A), which you know from the outset anyway, since you know that at low speeds it has to look like a nonrelativistic uniform acceleration, i.e. it must approach a linear relationship. So let's write it out in terms of the acceleration $a$; it is:

$$v=\frac{a\, t}{\sqrt{1+a^2\,\frac{t^2}{c^2}}}$$

This $a$ is the constant "gravity" that the spacefarer in the accelerating vehicle would feel: they'd feel exactly as though standing on Earth if $a=g$. Incidentally, it works out to almost exactly $2\,c/3$ if you accelerate at $1\,g$ for a year (Earth time).

This problem is treated in more detail in section 6.2 of Misner, Thorne and Wheeler, "Gravitation". John Baez explores this problem (although he too cites the same MTW source) in relation to our ability to explore space on his Relativistic Rocket Page. The Misner Thorne and Wheeler treatment is more general than mine, but the above is a simple, first principles reasoning. I'd advise you to look up and learn about four velocities and accelerations, if you've not already studies these.