This applet calculates a round trip to a star at a distance D from the origin (specified in light years). The fraction of time spent in accelerated motion, explained in this article, is f. Γ is the maximum Lorentz factor reached. The signals are sent at constant intervals of proper time (i.e. in the traveller’s non-inertial frame). The left panel is a space-time diagram, showing the time vs. displacement in the origin’s frame. The thick blue line is the spaceships’ path, and the thin red lines are the signals (travelling at the speed of light back to the origin). The right panel shows the time a signal is received (in the origin’s frame) as a function of the proper time when it was emitted.
In the summary table, V is the maximum velocity reached, a’ is the acceleration experienced (proper acceleration), Tproper is the length of proper time passed (the trip’s duration as experienced by the traveller), T is the length of time passed in the origin’s frame (the trip’s duration as experienced by a stationary observer at the origin), and γeff is just their ratio (that is the “effective” Lorentz factor of the trip).
The default values correspond to a trip to the nearest star system, Alpha Centauri. The trip has three phases of comfortable 10 m/s2 acceleration and two phases of coasting at a constant velocity of 60% the speed of light. The traveller is emitting signals at approximately one year intervals. If a twin undergoes this trip, they will find their sibling 2 years and 10 months older than they are upon their return to Earth. For the “classical” twin paradox, set f = 0.
Introduction
The formula for the displacement given constant acceleration in one dimension is a staple of high school physics:
\[x(t)=x_{0}+v_{0}t+{\textstyle \frac{1}{2}at^{2}}.\]where $a$ is the constant acceleration, and $x_{0}$ and $v_{0}$ crop up as integration constants. This formula stems from the definition of acceleration as the second derivative of position (or displacement) and in a sense remains “true” regardless of whether it describes plausible physical motion (i.e. at a sublight velocity) or not.
When I was taught Special Relativity, accelerated motion was politely swept under the rug. I was told that acceleration is within the realm of General Relativity, so we don’t touch it (although when I eventually took classes in General Relativity, we looked at free falling bodies rather than ones that accelerate under their own power). Special Relativity homework exercises focused on how measurements differ for different inertial observers through the Lorentz transformation (e.g. an observer on the Earth and another on a spaceship that travels at some constant speed $v$ with respect to the Earth).
Acceleration can however be explored within special relativity! An accelerating frame is not an inertial one, but at the proper time $\tau$ is still well defined, as well as the displacement $x$ and time $t$ as measured by some inertial observer. In this article I will show that the formulae equivalent to displacement given constant acceleration are:
$$ \begin{align*} x(\tau) & =x_{0}+\frac{c^{2}}{2a'\sqrt{k}}\left(1-e^{-a'\tau/c}\right)\left(e^{a'\tau/c}-k\right)\\ t(\tau) & =t_{0}+\frac{c}{2a'\sqrt{k}}\left(1-e^{-a'\tau/c}\right)\left(e^{a'\tau/c}+k\right) \end{align*} $$
where
$$k\equiv\frac{1-\beta_{0}}{1+\beta_{0}}=\frac{c-v_{0}}{c+v_{0}}$$
is related to the initial conditions (at proper time $\tau = 0$) and is similar to the Doppler factor.
Simple initial conditions
It actually surprised me that there is a reasonably elegant solution to this problem. If the initial conditions are simplified: at $\tau=0$ the traveller is instantaneously at rest at the origin, and the time in the origin’s frame is $t=0$, then
\[\begin{align*} x(\tau) &= \frac{c^{2}}{a'}\left[\cosh\left(a'\tau/c\right)-1\right]\\ t(\tau) &= \frac{c}{a'}\sinh\left(a'\tau/c\right) \end{align*}\]Note that at $\tau\gg c/a’$ both the hyperbolic sine and cosine as essentially exponential, so $x\sim t\sim e^{a’\tau/c}$. Similarly, the instantaneous velocity $v$ and Lorentz factor $\gamma$ are simple hyperbolic functions of $\tau$, finding them is left as an exercise for the reader.
Both $x$ and $t$ are expressed as functions of $\tau$; it is in fact possible to express $x$ as a function of $t$ but unless the initial conditions are very simple (as we are considering here), the expression is horrendously convoluted and not very informative. In the current situation though,
\[x(t)=\frac{c^{2}}{a'}\left[\sqrt{1+\left(a't/c\right)^{2}}-1\right].\]The asymptotic behaviour of this expression is:
\[x(t)= \begin{cases} {\textstyle \frac{1}{2}}a't^{2} & t\ll c/a'\\ -c^{2}/a'+ct & t\gg c/a' \end{cases}\]so at the very beginning of the accelerated motion we get the expected quadratic dependence. The acceleration at the Earth’s frame is virtually identical to the one in the spaceship’s instantaneous inertial frame, $a \approx a’$. After a some time, $x$ becomes linear with time, indicating that the traveller’s velocity at the origin’s frame is constant, and not surprisingly this constant velocity is the speed of light, the maximum speed possible.
Derivation
I’m going to show the basic idea behind deriving the expressions for $x(\tau)$ and $t(\tau)$. It’s based on having the traveller momentarily at rest at the origin of some prime coordinate system with known time offset, displacement, and velocity with respect to the stationary (non-prime) coordinate system (the Earth). Imagine the traveller experiencing some constant acceleration $a’$ in that prime coordinate frame for a very short interval of proper time $\Delta\tau$. At the end of that interval, the time in that frame is $t’=\Delta\tau$, the velocity is $u’=a’\Delta\tau$ and the position of the traveller is $x’=\frac{1}{2}a’\Delta\tau^{2}$ with respect to where it was in the beginning.
We can use the formula for addition of velocities in Special Relativity, and find the velocity of the traveller in the original frame:
\[u=\frac{v+u'}{1+\frac{vu'}{c^{2}}}\]Soon we will want to “reset” the prime coordinate frame such that the traveller is again at rest at the origin of this frame. First let us look at the change of the velocity in the non-prime coordinate system:
\[\begin{align*} \Delta v & =u-v\\ &= \frac{v+a'\Delta\tau}{1+\frac{va'\Delta\tau}{c^{2}}}-v\\ &= \frac{1-v^{2}/c^{2}}{1+va'\Delta\tau/c^{2}}a'\Delta\tau \end{align*}\]recognizing that the second term in the denominator is infinitesimal, we are left with
\[\Delta v=(1-v^{2}/c^{2})a'\Delta\tau\]or in other words, an ordinary differential equation (ODE):
\[\frac{\mathrm{d}v}{\mathrm{d}\tau}=(1-v^{2}/c^{2})a'.\]I won’t go through how to solve an ODE in general, but this one has a relatively simple solution. It is easy to verify that this is expression satisfies the equation:
\[v(\tau)=c\frac{e^{2a'\tau/c}-k}{e^{2a'\tau/c}+k}\]where $k$ is a constant, and in order to satisfy the initial conditions $v=v_{0}$ at $\tau=0$, it has to be as shown in the introduction.
Since the solution is not in terms of the time $t$, the displacement is not simply the integral. Instead we have to go to the Lorentz transformation:
\[\begin{align*} t &= t_{0}+\gamma(t'+vx'/c^{2})\\ x &= x_{0}+\gamma(x'+vt') \end{align*}\]substituting
\[\begin{align*} \Delta t &= \gamma\Delta\tau+{\textstyle \frac{1}{2}}\gamma va'/c^{2}\Delta\tau^{2}\\ \Delta x &= \gamma v\Delta\tau+{\textstyle \frac{1}{2}}\gamma a'\Delta\tau^{2} \end{align*}\]getting rid of quadratic terms, and the two equations are:
\[\begin{align*} \frac{\mathrm{d}t}{\mathrm{d}\tau} &= \gamma\\ \frac{\mathrm{d}x}{\mathrm{d}\tau} &= \gamma v \end{align*}\]The velocity $v$ is a now known function of $\tau$, and the Lorentz factor depends on $\tau$ through $v$ per its definition. By substituting $v(\tau)$ and integrating the right hand sides of both equations, and requiring $x(0)=x_0$ and $t(0)=t_0$, we get after a lot of algebra our relatively simple result.
Twin paradox
The twin paradox is the famous thought experiment where one twin travels to a different star and back at high velocity, and upon returning finds that their sibling that remained on Earth is now older than they are. The paradox is the apparent symmetry of the situation: from the travelling twin’s point of view, it is the Earth-bound twin who is moving at high velocity. So if that twin had only a very naïve understanding of Special Relativity, they might expect that their Earth-bound sibling would age more slowly and be younger rather than older when they reunite.
This is one of the problems that I was taught was “outside the scope” of Special Relativity, as the travelling twin experiences acceleration and thus a non-inertial observer. The only partly satisfactory solution I was taught was that the situation is not truly symmetric: only one of the twins experiences acceleration, and that breaks the symmetry. The paradox however can be easily understood well within Special Relativity, and even without considering acceleration at all (see the article on Wikipedia). However since this article is about acceleration, we can calculate $t(\tau)$ for the whole round trip. $t$ being the time measured by the Earth-bound twin, and $\tau$ the time measured by the travelling twin.
To make the problem interesting and also similar enough to the “classical” paradox, let us assume five phases of the trip: (1) acceleration from rest at the Earth to some maximum velocity $V$, corresponding to a Lorentz factor $\Gamma$; (2) coasting at this constant velocity; (3) decelerating at the same rate as the first phase, until we instantaneously stop and accelerate in the opposite direction until the velocity is again $V$; (4) again coasting at this velocity, now toward the Earth; (5) decelerating and stopping at the Earth.
Let us assume that the traveller powers their engine for a duration of $\Delta\tau_{\mathrm{acc}}$ in phases (1) and (5) and $2\Delta\tau_{\mathrm{acc}}$ in phase (3) (because they have to reverse direction). Phases (2) and (4) last for a duration of $\Delta\tau_{\mathrm{coast}}$ as experienced by the traveller. Since we fixed the maximum velocity, the proper acceleration $a’$ will just be expressed in terms of $V$ (or $\Gamma$) and $\Delta\tau_{\mathrm{acc}}$.
Phase 1
In this phase, the initial conditions are “simple” so we already saw the equations for displacement and time. We can find the velocity (in the Earth’s frame) by deriving $x(t)$ with respect to $t$, and then substitute $t(\tau)$ so the answer is expressed as a function of $\tau$. We get simply: \(v_{1}(\tau)=c\tanh(a'\tau/c).\) Now we isolate $a’$ by substituting $v_{1}(\Delta\tau_{\mathrm{acc}})=V$. That gives:
\[a'=(c/\Delta\tau_{\mathrm{acc}})\mathrm{arcosh}(\Gamma).\]So we can write:
\[\begin{align*} x_{1}(\tau) &= \frac{c\Delta\tau_{\mathrm{acc}}}{\mathrm{arcosh}(\Gamma)}\left\{ \cosh\left[\mathrm{arcosh}(\Gamma)\frac{\tau}{\Delta\tau_{\mathrm{acc}}}\right]-1\right\} \\ t_{1}(\tau) &= \frac{\Delta\tau_{\mathrm{acc}}}{\mathrm{arcosh}(\Gamma)}\sinh\left[\mathrm{arcosh}(\Gamma)\frac{\tau}{\Delta\tau_{\mathrm{acc}}}\right] \end{align*}\]The total distance travelled during this phase, and the time it took in the Earth’s frame are:
\[\begin{align*} \Delta x_{\mathrm{acc}}\equiv x_{1}(\Delta\tau_{\mathrm{acc}}) &= \frac{\left(\Gamma-1\right)}{\mathrm{arcosh}(\Gamma)}c\Delta\tau_{\mathrm{acc}}\\ \Delta t_{\mathrm{acc}}\equiv t_{1}(\Delta\tau_{\mathrm{acc}}) &= \frac{\sqrt{\Gamma^{2}-1}}{\mathrm{arcosh}(\Gamma)}\Delta\tau_{\mathrm{acc}} \end{align*}\]Phase 2
In the second phase, the coasting, we can’t use the previously derived formulae because we have the acceleration in the denominator which would result in division by zero. Instead, we just use the plain old Lorentz transformation:
\[\begin{align*} x_{2}(\tau) &= \Delta x_{\mathrm{acc}}+\Gamma V\tau_{2}\\ t_{2}(\tau) &= \Delta t_{\mathrm{acc}}+\Gamma\tau_{2} \end{align*}\]and for simplicity we define $\tau_{2}\equiv\tau-\Delta\tau_{\mathrm{acc}}$. So the distance travelled and time spent in this phase are:
\[\begin{align*} \Delta x_{\mathrm{coast}} &= \Gamma V\Delta\tau_{\mathrm{coast}}\\ \Delta t_{\mathrm{coast}} &= \Gamma\Delta\tau_{\mathrm{coast}} \end{align*}\]Phase 3
In this phase, we first decelerate to a halt and then accelerate in the opposite direction. The acceleration is constant, same as in phase 1 but with opposite signs. The initial conditions are not as simple now since this phase does not start from rest at the origin.
The $x_{0}$ and $t_{0}$ constants are where and when the spaceship is at the end of phase 2, that is $\Delta x_{\mathrm{acc}}+\Delta x_{\mathrm{coast}}$ and $\Delta t_{\mathrm{acc}}+\Delta t_{\mathrm{coast}}$, respectively. The $k$ constant is related to the coasting velocity:
\[k_{\mathrm{out}}=\frac{c-V}{c+V}.\]and we use the “out” subscript to denote this is related to the velocity at the outbound coasting phase. The expressions are therefore (don’t forget to flip the sign of $a’$; also we use the explicit definition of the inverse hyperbolic cosine to simplify):
\[\begin{align*} x_{3}(\tau) &= \Delta x_{\mathrm{acc}}+\Delta x_{\mathrm{coast}}-\frac{c\Delta\tau_{\mathrm{acc}}}{2\mathrm{arcosh}(\Gamma)\sqrt{k_{\mathrm{out}}}}\left(1-H^{\tau_{3}/\Delta\tau_{\mathrm{acc}}}\right)\left(H^{-\tau_{3}/\Delta\tau_{\mathrm{acc}}}-k_{\mathrm{out}}\right)\\ t_{3}(\tau) &= \Delta t_{\mathrm{acc}}+\Delta t_{\mathrm{coast}}-\frac{\Delta\tau_{\mathrm{acc}}}{2\mathrm{arcosh}(\Gamma)\sqrt{k_{\mathrm{out}}}}\left(1-H^{\tau_{3}/\Delta\tau_{\mathrm{acc}}}\right)\left(H^{-\tau_{3}/\Delta\tau_{\mathrm{acc}}}+k_{\mathrm{out}}\right) \end{align*}\]where we define:
\[\begin{align*} H &\equiv \Gamma+\sqrt{\Gamma^{2}-1}\\ \tau_{3} &\equiv \tau-\Delta\tau_{\mathrm{acc}}-\Delta\tau_{\mathrm{coast}} \end{align*}\]We can convince ourselves that $\Delta x$ and $\Delta t$ at both halves of this pase equal $\Delta x_{\mathrm{acc}}$ and $\Delta t_{\mathrm{acc}}$, respectively, as in phase 1. This has to be the case because of symmetry, but can also be shown by substituting $\Delta\tau_{\mathrm{acc}}$ and $2\Delta\tau_{\mathrm{acc}}$ in the expressions above and laboriously simplifying them.
The distance to the destination is thus:
\[D=2\Delta x_{\mathrm{acc}}+\Delta x_{\mathrm{coast}}\]Phase 4
At the beginning of this coasting phase, the spaceship now finds itself again at $x=\Delta x_{\mathrm{acc}}+\Delta x_{\mathrm{coast}}$, and the time elapsed on Earth it $t=3\Delta x_{\mathrm{acc}}+\Delta x_{\mathrm{coast}}$. The equations of motion are (not to forget to flip the sign of the velocity):
\[\begin{align*} t_{4}(\tau) & =3\Delta t_{\mathrm{acc}}+\Delta t_{\mathrm{coast}}+\Gamma\tau_{4}\\ x_{4}(\tau) &= \Delta x_{\mathrm{acc}}+\Delta x_{\mathrm{coast}}-\Gamma V\tau_{4} \end{align*}\]and again for simplicity we define $\tau_{4}\equiv\tau-3\Delta\tau_{\mathrm{acc}}-\Delta\tau_{\mathrm{coast}}$.
Phase 5
At the beginning of this coasting phase, the spaceship now finds itself again at $x=\Delta x_{\mathrm{acc}}$, and the time elapsed on Earth it $t=3\Delta x_{\mathrm{acc}}+2\Delta x_{\mathrm{coast}}$. The $k$ constant is different from phase 3 because the velocity at the beginning of the interval has the opposite sign. Therefore
\[k_{\mathrm{in}}=\frac{c+V}{c-V}=\frac{1}{k_{\mathrm{out}}}.\]The sign of acceleration is the same as phase 1, and finally we have:
\[\begin{align*} x_{5}(\tau) &= \Delta x_{\mathrm{acc}}+\frac{\sqrt{k_{\mathrm{out}}}c\Delta\tau_{\mathrm{acc}}}{2\mathrm{arcosh}(\Gamma)}\left(1-H^{-\tau_{5}/\Delta\tau_{\mathrm{acc}}}\right)\left(H^{\tau_{5}/\Delta\tau_{\mathrm{acc}}}-1/k_{\mathrm{out}}\right)\\ t_{5}(\tau) & =3\Delta t_{\mathrm{acc}}+2\Delta t_{\mathrm{coast}}+\frac{\sqrt{k_{\mathrm{out}}}\Delta\tau_{\mathrm{acc}}}{2\mathrm{arcosh}(\Gamma)}\left(1-H^{-\tau_{5}/\Delta\tau_{\mathrm{acc}}}\right)\left(H^{\tau_{5}/\Delta\tau_{\mathrm{acc}}}+1/k_{\mathrm{out}}\right) \end{align*}\]Summary
So now we know the full equation of motion for this trip and can create a nice plot as shown at the top of the page. But even without most of this algebra we could say that the time passing in the Earth-bound twin’s frame is always longer than the proper time for the travelling twin. In the case of the “classical” twin paradox, the ratio between the times is the Lorentz factor of the spaceship. In the case where there are acceleration and deceleration phases, the ratio is something else but still very simply dependent on the Lorentz factor, that is, the maximum Lorentz factor reached by the spaceship.