# How can acceleration in special relativity be uniform?

Acceleration is defied as the rate of change in velocity, implying that $$v(t) = at + v_0$$. Say that an object is accelerating at $$5 m/s^2$$ with respect to an inertial frame in special relativity. Using the definition of acceleration alone, we now have $$v(t) = 5t + v_0$$. This means that given enough time, $$v(t) > c$$ which is supposed to be impossible in relativity. The solution is to say that uniform acceleration in special relativity is hyperbolic, so while it may approach c it never actually gets there. The problem with this is how can that be said to be uniform acceleration? If something is accelerating at $$5m/s^2$$ in order for it to stay less than $$c$$ it has to eventually decelerate even more and more to asymptotically approach $$c$$.

How is hyperbolic acceleration uniform?

• In relativity, there are different ways to define acceleration. Your definition is that the acceleration is defined in a fixed reference frame. However, hyperbolic motion has uniform acceleration when considering proper acceleration (en.wikipedia.org/wiki/Proper_acceleration). The idea is to measure acceleration in the reference frame of the accelerating object (which changes), or geometrically by looking at the curvature of the word line in space-time.
– LPZ
Nov 13, 2022 at 15:00
• @lpz doesn’t the accelerating object see themselves as stationary?
– user310742
Nov 13, 2022 at 16:48
• Does this answer your question ? Nov 13, 2022 at 17:22
• To be precise: coordinate acceleration cannot be uniform. Proper acceleration can. The latter is measured by an accelerometer that the accelerated frame carries with itself. You should study the Rindler metric in detail. It is a beautiful piece of work. Nov 13, 2022 at 17:31
• Good followup from @KurtG. I'll just clarify my comment: your accelerated observer lies in a rest frame at every instant (corresponding to his instantaneous velocity). However, since he accelerates, he does not stay in this frame, call it $1$ but rather switches to a new one say $2$. Proper acceleration translates the acceleration needed to go from $1$ to $2$. Btw, a geometrical POV can also help. Hyperbolic motion is to Minkowski space-time as the circle is to the Euclidean plane (with proper acceleration being the analogue of curvature).
– LPZ
Nov 15, 2022 at 22:58

## 1 Answer

I’ll just formalize my previous comments. Let me restrict to 2D flat spacetime with a certain inertial frame $$t,x$$ ($$c=1$$ and the metric signature is $$(+,—)$$ like in particle physics). Then hyperbolic motion of proper acceleration $$a$$ can be parametrized by its proper time $$\tau$$ given (up to a space-time translation): $$t = \frac{\sinh(a\tau)}{a} \\ x = \frac{\cosh(a\tau)}{a} \\$$ As you pointed out, the acceleration viewed from the original frame is not uniform: $$x = \frac{\sqrt{1+(at)^2}}{a} \\ \frac{dx}{dt} = \frac{at}{\sqrt{1+(at)^2}}\\ \frac{d^2x}{dt^2} = -a\frac{1}{\sqrt{1+(at)^2}^3}$$ the velocity reaches $$1$$ asymptotically, so you have a non uniform deceleration reaching $$0$$ asymptotically.

Note however that acceleration $$d^2x/dt^2$$ in the frame coincides exactly with the proper acceleration $$a$$ at $$t=0$$ ie when the frame coincides with the rest frame of the particle.

This is true in general. At any event of the world-line, I can choose an inertial frame which coincides with the rest frame of the particle at that event. The acceleration measured in these identical frames at this specific event will coincide with proper acceleration. This is an equivalent definition of proper acceleration. And it is this proper acceleration that is constant in hyperbolic motion.

Geometrically, this instantaneous rest frame of the particle is the Minkowski analogue of the Frenet basis in the Euclicdean plane. This is why proper acceleration is the analogue of curvature and given generally by: $$a=\frac{d^2x}{d\tau^2}\frac{dt}{d\tau}-\frac{d^2t}{d\tau^2}\frac{dx}{d\tau}$$ which you can check explicitly for hyperbolic motion.

Hope this helps.

• Thanks for the answer. It makes a lot of sense. Just one more thing—you write the coordinates (x,t) as functions of proper time. If I’m understanding correctly, the proper time is the amount of time that the inertial observer measures. Therefore, if we want to look at the position of the accelerating object as a function of the time that the inertial observer sees, shouldn’t we look at $x(\tau)$ instead of $x(t)$? And hence $a(\tau)$?
– user310742
Dec 12, 2022 at 14:15
• Actually proper time is not the time measured by an inertial observer in general, as its name implies, it's the time the moving observer measures (if it happens to be an inertial observer, then you're right). In general when a frame $R$ has coordinate $x,t$, if you want to know whether a point accelerates with respects to $R$, you need to compute $d^2x/dt^2$ by definition.
– LPZ
Dec 12, 2022 at 17:14