# Proper Acceleration

The proper acceleration of an observer is that observer’s own sensed acceleration. Its description in terms of felt force would require an excursion into dynamics, but it can be defined kinematically if the observer is aware of some other frame which is known for some reason to be inertial. After acceleration of an observer, the previous frame of that observer appears (to that observer) to be moving in the opposite direction. So if the previous inertial frame (which matched the observer at time $t$ on the traveller’s own clock) is observed at time $t+\Delta t$ to have relative velocity $-\Delta v$, then the limiting ratio $a(t)=\lim\limits_{\Delta t\to 0}\frac{\Delta v}{\Delta t}$ is what we take as the proper acceleration. (We will ensure later that this purely kinematic definition does indeed match the dynamical definition in terms of felt force but for now just take it as “the” definition of proper acceleration.)

Any observer, $\mathcal{O}$, lives through a sequence of inertial frames $\mathcal{F}_{\mathcal{O}}(t)$, labelled by the “proper time” $t$ showing on $\mathcal{O}$’s clock, and each such frame associates any other frame with a relative velocity and a space-time displacement.

Let us use $v_{t}$, $t_{t}$, and $x_{t}$ to represent possible velocities and coordinates relative to the inertial frame $\mathcal{F}_{\mathcal{O}}(t)$, with $v_{t}(\mathcal{F})$, $t_{t}(\mathcal{F})$, and $x_{t}(\mathcal{F})$ being the values of those variables corresponding to some other frame $\mathcal{F}$ as seen from $\mathcal{F}_{\mathcal{O}}(t)$.

If $\mathcal{O}$ has proper acceleration $a(t)$, then the frame at time $t+\Delta t$ is thus seen from that at time $t$ as having velocity and coordinates given by $v_{t}=v_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=a(t)\Delta t +O((\Delta t)^{2})\approx a(t)\Delta t$, $x_{t}=x_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=a(t)\frac{(\Delta t)^{2}}{2}+O((\Delta t)^{3})\approx a(t)\frac{(\Delta t)^{2}}{2}$, and $t_{t}=t_{t}(\mathcal{F}_{\mathcal{O}}(t+\Delta t))=\Delta t +O((\Delta t)^{2})\approx \Delta t$ [where the reason this last is not exactly $\Delta t$ is because, as soon as the velocity changes, the clock measuring that $\Delta t$ for the accelerated observer is moving (and so slowed down) relative to that of the observer at time $t$].

[The inertial frame of the observer at time $t+\Delta t$ thus has time axis $x_{t+\Delta t}=0$ given in terms of the time $t$ coordinates approximately by $x_{t}-a(t)\frac{(\Delta t)^{2}}{2}=(a(t)\Delta t)(t_{t}-\Delta t)$, and the spacelike axis of simultaneity at $t_{t+\Delta t}=0$ is given by $t_{t}-\Delta t=(a(t)\Delta t)(x_{t}-a(t)\frac{(\Delta t)^{2}}{2})$.

So the simultaneity space $t_{t+\Delta t}=0$ intersects $t_{t}=0$ at the event where $-\Delta t=(a(t)\Delta t)(x_{t}-a(t)\frac{(\Delta t)^{2}}{2})$, i.e. $x_{t}=-\frac{1}{a(t)}+O((\Delta t)^2)$.

For the case of constant acceleration $a$ this makes all the simultaneity spaces pass through the same “fulcrum event” and this can be used to get a geometric proof that the worldline is hyperbolic and that events beyond the “Rindler horizon” (which is always at $x_{t}=-\frac{1}{a}$) remain forever inaccessible to the accelerated observer.

This approach is discussed in more detail in the Wikipedia page on Rindler Coordinates (see also this version on the anonymously authored ‘mathpages’ website, and this discussion by Greg Egan). But in the discussion below I shall take what I think is a more intuitively direct path, and try to say something about the case of non-constant proper acceleration as well.]

As judged by the “birth” frame (corresponding to $t=0$), a difference of $\Delta t$ in the value of $t_{t}(\mathcal{F})$ corresponds to a difference of $\gamma(t)\Delta t$ in $t_{0}(\mathcal{F})$(where $\gamma(t)=\frac{1}{\sqrt{1-{v_{b}(t)}^{2}}}$).

[Note: It is also true that, as judged by the frame at time $t$, a difference of $\Delta t$ in the value of $t_{0}(\mathcal{F})$ corresponds to a difference of $\gamma(t)\Delta t$ in $t_{t}(\mathcal{F})$ (since the question of which frame appears to be time dilated depends on which frame is making the comparison). And if this still seems confusing then you may need to re-visit the earlier discussion of time dilation.]

So, if we use $t_{b}(t)$ as shorthand for $t_{0}(\mathcal{F}_{\mathcal{O}}(t))$, and similarly for $x_{b}(t)$ and $v_{b}(t)$, then the relations above can be written in terms of the birthframe coordinates as follows:

$t_{b}(t+\Delta t)=t_{b}(t)+\gamma(t) \Delta t+O((\Delta t)^{2})$,

$x_{b}(t+\Delta t)=x_{b}(t)+v_{b}(t)(\gamma(t) \Delta t)+O((\Delta t)^{2})$, and

\begin{align}v_{b}(t+\Delta t)&=v_{b}(t)[+]a(t)\Delta t =\frac{v_{b}(t)+a(t)\Delta t}{1+(v_{b}(t))(a(t)\Delta t)}\\&=(v_{b}(t)+a(t)\Delta t)(1 – (v_{b}(t))(a(t)\Delta t))+O((\Delta t)^{2})\\&= v_{b}(t)+a(t)\Delta t – (v_{b}(t))^{2}(a(t)\Delta t)+O((\Delta t)^{2})\end{align}

(where [+] is just shorthand for the relativistic velocity sum).

So we have the differential equations

$t_{b}'(t)=\gamma(t)=\frac{1}{\sqrt{1-{v_{b}(t)}^{2}}}$,

$x_{b}'(t)=\gamma(t)v_{b}(t)=\frac{v_{b}(t)}{\sqrt{1-{v_{b}(t)}^{2}}}$, and

$v_{b}'(t)=a(t)(1-{v_{b}(t)}^{2})$.

The last of these is separable and so is easily solved (by cross-multiplying and integrating) to get $\int \frac{v_{b}'(t)dt}{(1-{v_{b}(t)}^{2})}=\int a(t)dt$.

This then gives $\ln{\sqrt{\frac{1+v}{1-v}}}=\int a(t)dt$, which gives $\frac{1+v}{1-v}=e^{2\int a(t)dt}$ and so $v_{b}(t)=\frac{e^{2\int a(t)dt}-1}{e^{2\int a(t)dt}+1}=\tanh(\int a(t)dt)$.

Note: Here the integrated proper acceleration $\int a(t)dt$ is the subjective velocity change, which from now on we’ll call $v_{s}(0,t)$, and $v_{b}(t)=\tanh(v_{s}(0,t))$ is the apparent velocity of the traveller as inferred by the birthframe observer at the birthframe time when the traveller’s clock reads time $t$.

For the coordinates, with respect to the birthframe, of $\mathcal{O}$’s worldline event at proper time $t$ we now get

\begin{align}t_{b}(t)&=\int\frac{dt}{\sqrt{1-{v_{b}(t)}^{2}}}=\int\frac{dt}{\sqrt{1-{\tanh(v_{s}(0,t)}^{2}}}\\&=\int\cosh(v_{s}(0,t))dt\end{align}

and

\begin{align}x_{b}(t)&=\int\frac{v_{b}(t)dt}{\sqrt{1-{v_{b}(t)}^{2}}}=\int\cosh(v_{s}(0,t))\tanh(v_{s}(0,t))dt\\&=\int\sinh(v_{s}(0,t))dt\end{align}.

Note: By FTC&ChainRule, $x_{b}'(t)=\sinh(v_{s}(0,t)dt)\frac{d}{dt}v_{s}(0,t)$ and $t_{b}'(t)=\cosh(v_{s}(0,t)dt)\frac{d}{dt}v_{s}(0,t)$, so the apparent velocity as seen by the birthframe observer is given by $v_{b}=\frac{dx_{b}}{dt_{b}}=\frac{x_{b}'(t)}{t_{b}'(t)}=\tanh(v_{s}(0,t))$.

This is in agreement with our previously derived formula for $v_{b}(t)$, but to get it in terms of the $t_{b}$ coordinate we need to compose with the function giving the proper time $t$ of $\mathcal{O}$ when it is at birthframe time $t_{b}$ – which is the composition inverse of the function $t_{b}(t)=\int\cosh(\textstyle\int a(t)dt)dt=\int\cosh(v_{s}(0,t))dt$ .

In general, the above integral is not easy to compute explicitly, but in the special case that $a(t)=a$ is constant, then we have $v_{s}(0,t)=at+C_{v}$, so $v_{b}(t)=\tanh(at+C_{v})$ (with $v_{b}(0)=0$ giving $C_{v}=0$), and in that case we can complete the integrals for the coordinates to get

$t_{b}(t)=\int\cosh(at+C_{v})dt=\frac{1}{a}\sinh(at+C_{v})+C_{t}$ with $t_{b}(0)=0$ giving $C_{t}=0$, so $t_{b}(t)=\frac{1}{a}\sinh(at)$, and

$x_{b}(t)=\int\sinh(at+C_{v})dt=\frac{1}{a}\cosh(at+C_{v})+C_{x}$ with $x_{b}(0)=0$ giving $C_{x}=-\frac{1}{a}$, so $x_{b}(t)=\frac{1}{a}\cosh(at)-\frac{1}{a}$.

And in this case we can invert to get $t(t_{b})=\frac{1}{a}[\sinh^{-1}(at_{b})]$ which gives

\begin{align}v_{b}(t_{b})&=\tanh(a[\frac{1}{a}[\sinh^{-1}(at_{b})])\\&=\tanh(\sinh^{-1}(at_{b}))=\frac{at_{b}}{\sqrt{1+(at_b)^2}}\end{align}

Which, as expected, remains less than $c$ (which is 1 in our units) and so the acceleration observed by the birthframe observer is decidedly NOT constant but rather decreasing to zero as the speed gets closer and closer to $c$.

[insert brief discussion of Rindler coords and horizon for the case of const accel for all time]

For the case of piecewise constant proper acceleration, say $a(t)=a_{i}$ between proper times $t_{i}$ and $t_{i+1}$, we can use the foregoing analysis to follow how the frame changes in each interval. But in this case the period of constant acceleration is starting at $t=t_{i}$ rather than $t=0$, so the conditions for the constants of integration (which depend on $i$) are now as follows:

$\tanh(a_{i}t_{i}+C_{v,i})=v_{b}(t_{i})$ gives $C_{v,i}=\tanh^{-1}(v_{b}(t_{i}))-a_{i}t_{i}=v_{s}(0,t_{i})-a_{i}t_{i}$, but we’ll often just use $(a_{i}t_{i}+C_{v,i})=\tanh^{-1}(v_{b}(t_{i}))=v_{s}(0,t_{i})$ directly;

$\frac{1}{a_{i}}\sinh(a_{i}t_{i}+C_{v,i})+C_{t,i}=t_{b}(t_{i})$ gives \begin{align}C_{t,i}&=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(a_{i}t_{i}+C_{v,i})=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(v_{s}(0,t_{i}))\\ \text{[also}&=t_{b}(t_{i})-\frac{1}{a_{i}}\sinh(\tanh^{-1}(v_{b}(t_{i})))\\&=t_{b}(t_{i})-\frac{1}{a_{i}}v_{b}(t_{i})\sqrt{1+v_{b}(t_{i})^2}\;\; \text{];}\\ \text{and}\end{align}

$\frac{1}{a_{i}}\cosh(a_{i}t_{i}+C_{v,i})+C_{x,i}=x_{b}(t_{i})$ gives \begin{align}C_{x,i}&=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(a_{i}t_{i}+C_{v,i})=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(v_{s}(0,t_{i}))\\ \text{[also}&=x_{b}(t_{i})-\frac{1}{a_{i}}\cosh(\tanh^{-1}(v_{b}(t_{i})))\\&=x_{b}(t_{i})-\frac{1}{a_{i}}\frac{v_{b}(t_{i})}{\sqrt{1+v_{b}(t_{i})^2}}\;\; \text{].}\end{align}

So, for $t_{i}<t<t_{i+1}$ we have

\begin{align}v_{b}(t)&=\tanh[\tanh^{-1}(v_{b}(t_{i}))+a_{i}(t-t_{i})]\\&=\frac{v_{b}(t_{i})+\tanh(a_{i}(t-t_{i}))}{1+(v_{b}(t_{i}))(\tanh(a_{i}(t-t_{i})))}=v_{b}(t_{i})[+]\tanh(a_{i}(t-t_{i}))\end{align}

with

\begin{align}t_{b}(t)&=\int\cosh[a_{i}t+C_{v,i}]dt=\frac{1}{a_{i}}\sinh[a_{i}t+C_{v,i}]+C_{t,i}\\&=\frac{1}{a_{i}}\sinh[a_{i}t+v_{s}(0,t_{i})-a_{i}t_{i}]+\{t_{b}(t_{i})-\frac{1}{a_{i}}\sinh[v_{s}(0,t_{i})]\}\\&=t_{b}(t_{i})+\frac{1}{a_{i}}\{\sinh[a_{i}(t-t_{i})+v_{s}(0,t_{i})]-\sinh[v_{s}(0,t_{i})]\}\text{,}\end{align}

(giving

\begin{align}t_{b}(t_{i+1})-t_{b}(t_{i})&=\frac{1}{a_{i}}\{\sinh[a_{i}(t_{i+1}-t_{i})+v_{s}(0,t_{i})]-\sinh[v_{s}(0,t_{i})]\}\\&=\frac{1}{a_{i}}\{\sinh[v_{s}(0,t_{i+1})]-\sinh[v_{s}(0,t_{i})]\}\\&=\int_{t_{i}}^{t_{i+1}}\cosh(v_{s}(0,t))dt\end{align} since $\frac{d}{dt}\sinh[v_{s}(0,t)]=\cosh[v_{s}(0,t)]\frac{d}{dt}v_{s}(0,t)$ and $\frac{d}{dt}v_{s}(0,t)=a_{i}$ for $t_{i}<t<t_{i+1}$ ),

and

\begin{align}x_{b}(t)&=\int\sinh[a_{i}t+C_{v,i}]dt=\frac{1}{a_{i}}\cosh[a_{i}t+C_{v,i}]+C_{x,i}\\&=\frac{1}{a_{i}}\cosh[a_{i}t+v_{s}(0,t_{i})-a_{i}t_{i}]+\{x_{b}(t_{i})-\frac{1}{a_{i}}\cosh[v_{s}(0,t_{i})]\}\\&=x_{b}(t_{i})+\frac{1}{a_{i}}\{\cosh[a_{i}(t-t_{i})+v_{s}(0,t_{i})]-\cosh[v_{s}(0,t_{i})]\}\text{.}\end{align}

(giving

\begin{align}x_{b}(t_{i+1})-x_{b}(t_{i})&=\frac{1}{a_{i}}\{\cosh[a_{i}(t_{i+1}-t_{i})+v_{s}(0,t_{i})]-\cosh[v_{s}(0,t_{i})]\}\\&=\frac{1}{a_{i}}\{\cosh[v_{s}(0,t_{i+1})]-\cosh[v_{s}(0,t_{i})]\}\\&=\int_{t_{i}}^{t_{i+1}}\sinh(v_{s}(0,t))dt\end{align} since $\frac{d}{dt}\cosh[v_{s}(0,t)]=\sinh[v_{s}(0,t)]\frac{d}{dt}v_{s}(0,t)$ and $\frac{d}{dt}v_{s}(0,t)=a_{i}$ for $t_{i}<t<t_{i+1}$).